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3 years ago

Evaluate the expression 3.14(a2 (a squared) + ab) when a = 4 and b = 2. (Input decimals only, such as 12.71, as the answer.)

1 answer:

7 0

First, start out by inserting the numbers:

3.14(4(2) (4^2) + 4(2)

Next, multiply the given numbers:

3.14(8 (4^2) + 8)

Then, multiply 4^2 (since the exponent is 2, multiply 4 two times: 4 * 4):

3.14(8 (16) + 8)

Now, multiply the first 8 to 16:

3.14(128 + 8)

Use the distributive property to solve this next one:

401.92 + 25.12

Finally, add up you two decimals:

427.04
You should get 427.04 (I could be wrong though).
Hopefully that helped! :)

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I’m confused on this one

andrew-mc [135]

First we need to know the point slope form. It is y - y1 = m(x - x1). In the y1 and x1 spots, you just plug in any coordinate on the line.

m would be the slope which we already know as 5/3.

We are given the coordinate point (0,3) as a point on the line. This is also what we are going to plug into the equation. 0 will be x1 and 3 will be y1.

Our equation should then look like after plugged in as....

y - 3 = 5/3(x)

x is by itself because subtracting or adding by 0 yield the original number.

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3 years ago

Choose the term that identifies the designated portion of this quadratic trinomial

Lana71 [14]

You did not give us the equation so we can't answer that. Update it and I will try to help.

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3 years ago

Oxana [17]

Answer:

Step-by-step explanation:

$a(n) = a1 *(1.25)^{(n-1)}$

$a(1) = 5$

$a(2) = 5*(1.25)^{(2-1)}\\a(2) = 5*(1.25)^{1}\\a(2) = 5*(1.25)\\a(2) = 6.25$

$a(3) = 5 * (1.25)^{(3-1)} \\a(3) = 5 * (1.25)^{2}\\a(3) = 5 * 1.565\\a(3) = 7.813$

$a(4) = 5* (1.25)^{(4-1)}\\ a(4) = 5* (1.25)^{3}\\a(4) = 5* 1.953\\a(4) = 9.766$

$a(5) = 5 * (1.25)^{(5-1)}\\a(5) = 5 * (1.25)^{4}\\a(5) = 5 * 2.441\\a(5) = 12.207$

$a(6) = 5 * (1.25)^{5}\\a(6) = 5 * 3.052\\a(6) = 5 * 3.0517\\a(6) = 15.259$

$a(1) = 5$

$a(2) = 6.25$

$a(3) = 7.813$

$a(4) = 9.766$

$a(5)=12.207$

$a(6)=15.259$

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3 years ago

line segment connecting the vertices of a hyperbola is called the ________ ________, and the midpoint of the line segment is the

Lesechka [4]

line segment connecting the vertices of a hyperbola is called the <u>transverse axis</u> and the midpoint of the line segment is the <u>center</u> of the hyperbola.

What is transverse axis and center of hyperbola ?

The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints.

And The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes.

Learn more about the transverse axis and center of hyperbola here:

brainly.com/question/28049753

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2 years ago

Three out of every ten dentists recommend a certain brand of fluoride toothpaste. Which assignment of random digits would be use

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Answer:

eddfdgdccggģdffcdrrfxddxcvgfx

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3 years ago