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3 years ago

13

The area of an artist's square canvas can hold 113 square inches of paint. What is the approximate length of one side of the can

vas? (Approximate to the nearest hundredth inch.)

2 answers:

Naya [18.7K]3 years ago

8 0

Area of canvas= 113 square in
area of one side= 113/4=28.22
=28 square in

fomenos3 years ago

3 0

it would be 10.63 inches

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50 points i need to know how to plot these

svetlana [45]

x is time and y is pencils that you have

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so at x=1, do 90-15=75 pencils

at x=2, do 75-15=60 pencils

etc

until you get to

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3 years ago

you rent an apartment that costs $1600 per month during the first year ,but the rent is set to go up 11.5% per year.what would b

RoseWind [281]

Answer:

$4,750.

Step-by-step explanation:

To determine the rental value of the apartment in the tenth year, knowing that it is worth 1,600 during the first year but that it will be adjusted annually to 11.5%, the following calculation must be made:

X = 1,600 x (1 + 0.115) ^ 10

X = 4,751.91

Thus, at ten years, rounded to the nearest tenth, the rental value will be $ 4,750.

7 0

3 years ago

Read 2 more answers

Which of the following is an equivalent equation obtained by completing the square of the expression below? x^2+6x−8=0 A (x+3)^2

valentina_108 [34]

Answer:

Solving the equation $x^2+6x-8=0$ by completing the square method we get $\mathbf{(x+3)^2=17}$

Option B is correct option.

Step-by-step explanation:

We need to solve the equation $x^2+6x-8=0$ by completing the square method.

For completing the square method: we need to follow: $a^2+2ab+b^2=(a+b)^2$

We are given:

$x^2+6x-8=0$

Solving by completing the square

$x^2+2(x)(?)+(?)^2-8=0$

We need to find ? in our case ? is 3 because 2*3= 6 and our middle term is 6x i,e 2(x)(3)=6x.

So, adding and subtracting (3)^2

$x^2+2(x)(3)+(3)^2-8-(3)^2=0\\(x+3)^2-8-9=0\\(x+3)^2-17=0\\(x+3)^2=17$

So, solving the equation $x^2+6x-8=0$ by completing the square method we get $\mathbf{(x+3)^2=17}$

Option B is correct option.

7 0

3 years ago

What is the value of p in the linear equation 24p + 12 – 18p = 10 + 2p – 6?

stira [4]

we have

$24p+12-18p=10+2p-6$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$24p-18p-2p=10-6-12$

Combine like terms

$4p=-8$

Divide by $4$ both sides

$4p/4=-8/4$

$p=-2$

therefore

<u>the answer is</u>

the value of p in the linear equation is $-2$

6 0

3 years ago

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Given that f(–2.4) = -1 and f(-1.9) = -8, approximate<br> f'(-2.4).<br> f'(-2.4)

lora16 [44]

Answer:

f'(-2.4) ≈ -14

General Formulas and Concepts:
<u>Algebra I</u>

Coordinate Planes

Coordinates (x, y)

Slope Formula: $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

Functions

Function Notation

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Step-by-step explanation:

*Note:

The definition of a derivative is the slope of the <em>tangent</em> <em>line</em>.

<u>Step 1: Define</u>

<em>Identify.</em>

f(-2.4) = -1

f(-1.9) = -8

<u>Step 2: Differentiate</u>

Simply plug in the 2 coordinates into the slope formula to find slope <em>m</em>.

[Derivative] Set up [Slope Formula]: $\displaystyle f'(-2.4) \approx \frac{f(x_2) - f(x_1)}{x_2 - x_1}$
Substitute in coordinates: $\displaystyle f'(-2.4) \approx \frac{-8 - -1}{-1.9 - -2.4}$
Evaluate: $\displaystyle f'(-2.4) \approx -14$

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Learn more about derivatives: brainly.com/question/17830594

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

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2 years ago