Find the value of:
7- (1/2)3​

f(x) < g(x)

(2x + 3) < (3x - 2)

3 < x - 2

5 < x

x > 5

Check:

2x + 3 < 3x - 2

2(5) + 3 < 3(5) - 2

10 + 3 < 15 - 2

13 < 13 --- FALSE

2(10) + 3 < 3(10) - 2

20 + 3 < 30 - 2

23 < 28 --- TRUE

Hope this helps!

Answer:

x < 5

Step-by-step explanation:

f(x) < g(x) if f(x) = 2x + 3 and g(x)= 3x - 2 ​

2x + 3 < 3x - 2 ​

2x - 3x < -2 -3

-x < -5

x > 5

9514 1404 393

Answer:

  domain: (-4, 9)

Step-by-step explanation:

The graph is defined for all points between -4 and 9, but not for those two points. In interval notation, the domain is (-4, 9).

1 st picture,

Other names of <1 are - <F, <YFP, <PFY

2 nd picture,

Acute

3 rd picture,

Point E

Answer:

1. ∠F, ∠YFP, ∠PFY

2. Acute

3. Point E

Step-by-step explanation:

Answer:

[tex]ab + 2 - 1 = ab + 1 \\ a = 3 \\ b = 4 \\ then \: 3 \times 4 + 1 \\ = 12 + 1 = 13 \\ thank \: you[/tex]

Answer:

so you basicly divided 5.87 to 3.25

Step-by-step explanation:

do this : 5.87/3.25=  your answer

It looks like the differential equation is

[tex]y'-\dfrac y{x+1}=2[/tex]

Multiply both sides by 1/(x + 1) :

[tex]\dfrac{y'}{x+1}-\dfrac y{(x+1)^2} = \dfrac2{x+1}[/tex]

The left side is now a derivative of a product,

[tex]\left(\dfrac y{x+1}\right)' = \dfrac2{x+1}[/tex]

Integrate both sides with respect to x :

[tex]\dfrac y{x+1} = 2\ln|x+1|+C[/tex]

Solve for y :

[tex]y = 2(x+1)\ln|x+1| + C(x+1)[/tex]

Answer:

SOH-CAH-TOA

Tan(angle) = Opposite side length/adjacent side lenght

Tan(49 deg) = h/900

Multiply both sides by 900:

900* tan(49deg) = h

Throw it into a calculator

h ≈ 1035.331566

Step-by-step explanation:

Answer:

x°=43°

2x°=86°

(x+8)°=51°

Step-by-step explanation:

x+2x+x+8=180°[being sum of the angles of triangle]

or, 4x+8=180

or, 4x=180-8

or, 4x=172

or, x=172/4

or, x=43°

Then,

2x=2×43°=86°

(x+8)°=(43+8)°=51°

Please mark me brainliest.

Answer:

A. 0.4

Step-by-step explanation:

Write 12 of 30 as a fraction:

12/30

Convert this to be a fraction over 100 by setting up an equation:

Variable x = proportional answer to 12/30

12/30 = x/100

Cross multiply:

12 × 100 = 30 × x

1200 = 30x

40 = x

40/100

Convert this number into a decimal by adding a zero and decimal point before the "4" in 40:

0.40

Simplify:

0.40 = 0.4

I converted it into a decimal through the use of fractions, but you can also find the answer by doing long division: 12 ÷ 30. However, showing work for long division over a message is too complex so I solved the problem this way instead.

Answer:

f(t)=-16t^2+8t+2

Step-by-step explanation:

Take the initial velocity(8)put in for vt. Then take height(2) and put in for s. giving you the equation. f(t)= -16t^2+8t+2

The answer you are looking for is DXCVI.

D=500.

X=10.

C=100.

V=5.

I=1

Solution:

So,

500+(100-10)+5+1

500+90+5+1

596

Therefore, DXCVI=596.

Answer:

596 in Roman Numerals should be DXCVI

Answer:

4x + 6 = 94 could be used to solve the problem

Answer:

about 41 cents

Step-by-step explanation:

Answer:

The equation in y = mx+b form is  y = (4/3)x - 20/3

The equation in Ax+By = C form is  4x - 3y = 20

There are other ways to express the second equation.

==============================================================

Explanation:

First we must complete the square for x and y.

This will get the equation into (x-h)^2+(y-k)^2 = r^2 form so we can determine the center.

x^2+y^2 - 8x-14y + 40 = 0

(x^2-8x) + (y^2-14y) + 40 = 0

(x^2-8x+16 - 16) + (y^2-14y+49-49) + 40 = 0

(x^2-8x+16) + (y^2-14y+49) + (-16-49+40) = 0

(x-4)^2 + (y-7)^2 = 25

(x-4)^2 + (y-7)^2 = 5^2

This circle is centered at (h,k) = (4,7) and it has radius r = 5.

----------------------

Next, we need to find the slope of line AB such that

A = (4,7) = center of the circle

B = (8,4) = point on the circle, which is the point of tangency

m = slope

m = (y2-y1)/(x2-x1)

m = (4-7)/(8-4)

m = -3/4

The slope of line AB is -3/4. Any perpendicular slope will have us do two things:

that would get us from -3/4 to +4/3 or just 4/3.

The perpendicular slope is 4/3.

----------------------

So we must find the equation of the line with slope m = 4/3 and it goes through (x,y) = (8,4)

That means,

y = mx+b

4 = (4/3)*8 + b

4 = 32/3 + b

4 - 32/3 = b

12/3 - 32/3 = b

(12-32)/3 = b

-20/3 = b

b = -20/3

Therefore, the equation of the tangent line is y = (4/3)x - 20/3

Optionally, we can get this equation into standard form like so

y = (4/3)x - 20/3

3y = 4x - 20 ... multiply everything by 3

3y-4x = -20

-4x+3y = -20

4x - 3y = 20

The diagram is shown below.

Answer:

?

Step-by-step explanation:

So first, you need to multiply 7 by four to get the quarterly part of it, which will be 28. Then, multiply 9% (.9) by your 28 to get 25.2. Finally, multiply that by 7,000 to get 176,400...

Answer:

Z, K and R

and

H ,B and C

........

Step-by-step explanation:

P(x,y)=(1,-3)

line is perpendicular to x=2 which shows that line is parallel to x-axis.

angle=0°

m=tan(angle)

m=tan0°

m=0

By using point slope form:

y+3=0(x-1)

y+3=0

Answer:

the answer will be 60 percent

=60%

Answer: y + 5 = 3/5 (x + 5)

Concept:

There are four general forms of equations:  

Here, we are going to use the point-slope form because the question has given us both the slope and one point.

Solve:

**Disclaimer** Since the question did not clarify, thus, I choose to use the point-slope form as it is the easiest method. However, if the question has a specified form of the equation, please tell me and I will redo it.

Given information

(x₁, y₁) = (-5, -5)

m = 3/5

Given expression

y - y₁ = m (x - x₁)

Substitute values into the expression

y + 5 = 3/5 (x + 5)

Hope this helps!! :)

Please let me know if you have any questions

Answer:

Step-by-step explanation:

New Balance = 110 + 110*2/100

New Balance = 110 + 2.20

New Balance = 112.20

The copy is the exact same as the original, so the scale factor would be 1.

The answer is the first choice

Answer:

[tex]4m + 5 - 3m \\ = 4m - 3m + 5 \\ = 1m + 5 \\ = (m + 5)[/tex]

Given the expression:

[tex] \displaystyle \large{4m + 5 - 3m}[/tex]

To simplify the expression, notice that 4m and 3m have the same term.

These are called like terms. Let's see some examples of like terms.

Then we also have the example of non-like term

In conclusion,

And that brings us to this topic, to evaluate or simplify the expression with two like terms and one non-like term.

Note that only like terms can evaluate. Let's see some examples:

How about a subtraction?

But what if it's non-like terms? Simple, we keep it like that!

So to simplify the expression in the question, we evaluate like terms first!

Since 4m and 3m both are like terms. We can do 4m-3m.

[tex] \displaystyle \large{4m + 5 - 3m} \\ \displaystyle \large{4m - 3m + 5} [/tex]

Evaluate 4m-3m

[tex] \displaystyle \large{1m + 5} [/tex]

We don't usually write 1m so we keep as m instead.

[tex] \displaystyle \large{m + 5} [/tex]

Now, can we evaluate m+5? No, because m and 5 are not like terms so we keep like this.

Hence, the answer is m+5

Answer:

550 mi. !!!!

Step-by-step explanation:

Answer: 550 miles

Work Shown:

1 in = 100 mi

5.5*(1 in) = 5.5*(100 mi) ..... multiply both sides by 5.5

5.5 in = 550 mi

Step-by-step explanation:

The quotient of three and number n translate to 3 divide by a number n,given the quotient indicates division or fraction. four more than that value tells us that we simply add a 4

There are some symbols missing in your integral. I suspect you meant something along the lines of

[tex]\displaystyle \int_C (x+5y)\,\mathrm dx + x^2\,\mathrm dy[/tex]

but we can consider a more general integral,

[tex]\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy[/tex]

One way to compute the integral is to split up C into two component paths C₁ and C₂, parameterize both, directly compute the integral over each path, then sum the results.

Parameterize C₁ and C₂ respectively by

〈x₁(t), y₁(t)⟩ = (1 - t ) 〈0, 0⟩ + t 〈5, 1⟩ = 〈5t, t⟩

〈x₂(t), y₂(t)⟩ = (1 - t ) 〈5, 1⟩ + t 〈5, 0⟩ = 〈5, 1 - t⟩

where 0 ≤ t ≤ 1. Then

[tex]\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy = \int_{C_1} f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy + \int_{C_2} f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy \\\\ = \int_{C_1} \left(f(x_1(t),y_1(t))\frac{\mathrm dx_1}{\mathrm dt} + g(x_1(t),y_1(t))\frac{\mathrm dy_1}{\mathrm dt}\right)\,\mathrm dt \\ + \int_{C_2} \left(f(x_2(t),y_2(t))\frac{\mathrm dx_2}{\mathrm dt} + g(x_2(t),y_2(t))\frac{\mathrm dy_2}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^1 \left(5f(5t,t) + g(5t,t) - g(5,1-t)[/tex]

If f and g agree with what I suggested earlier, the integral reduces to

[tex]\displaystyle \int_0^1 \left(5(5t+5t) + (5t)^2 - 5^2\right)\,\mathrm dt = \int_0^1 \left(25t^2 + 50t - 25\right)\,\mathrm dt = \boxed{\frac{25}3}[/tex]

Another way would be to close the path with a line segment from (5, 0) to (0, 0) and apply Green's theorem. Compute the resulting double integral, then subtract the contribution of the line integral over this third line segment. Provided that f(x,y) and g(x,y) don't have any singularities along this closed path C* or inside the triangle (call it T ) that it surrounds, we have

[tex]\displaystyle \int_{C^*} f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy = -\iint_T \frac{\partial g}{\partial x}-\frac{\partial f}{\partial y} \,\mathrm dx\,\mathrm dy[/tex]

where T is the region,

[tex]T = \left\{(x,y) \mid 0\le x\le 5 \text{ and } 0\le y\le \dfrac x5\right\}[/tex]

The double integral has a negative sign because C* has a negative, clockwise orientation.

Then

[tex]\displaystyle \int_{C^*} f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy = -\int_0^5\int_0^{x/5} \frac{\partial g}{\partial x}-\frac{\partial f}{\partial y} \,\mathrm dy\,\mathrm dx[/tex]

If f and g are as I suggested, then

[tex]\displaystyle \int_{C^*} f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy = \int_0^5\int_0^{x/5} (5-2x) \,\mathrm dy\,\mathrm dx = -\frac{25}6[/tex]

From this, we subtract the integral along the line segment C₃ from (5, 0) to (0, 0), parameterized by

〈x₃(t), y₃(t)⟩ = (1 - t ) 〈5, 0⟩ + t 〈0, 0⟩ = 〈5 - 5t, 0⟩

again with 0 ≤ t ≤ 1.

Then the remaining line integral is

[tex]\displaystyle \int_{C_3} f(x_3(t),y_3(t))\,\mathrm dx_3 + g(x_3(t),y_3(t))\,\mathrm dy_3 = \int_{C_3} x_3(t)\frac{\mathrm dx_3}{\mathrm dt}\,\mathrm dt \\\\ = \int_0^1 -5(5-5t)\,\mathrm dt = -\frac{25}2[/tex]

and so the original line integral is again -25/6 - (-25/2) = 25/3.