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3 years ago

HELP me on this please

Mathematics

1 answer:

Rzqust [24]3 years ago

6 0

Answer:

angle FEA is 100°

angle AEC is 20°

angle CEG is 60°

Step-by-step explanation:

angle AEC=angle DEB=20°

angle AEG=angle FED=80°

80°-20°=60°; angle CEG=60°

angle FEA+angle AEC+ angle CEG=180°

angle FEA=100°

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Y=x^2+bx+c;(-2,3)

Maksim231197 [3]

<h3>Answer: b = 4 and c = 7.</h3>

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

Explanation:

Comparing y = x^2+bx+c to y = ax^2+bx+c, we see that a = 1.

The vertex given is (-2,3). In general, the vertex is (h,k). So h = -2 and k = 3.

Plug those three values into the vertex form below

y = a(x-h)^2 + k

y = 1(x-(-2))^2 + 3

y = (x+2)^2 + 3

Then expand everything out and simplify

y = x^2+4x+4 + 3

y = x^2+4x+7

We see that b = 4 and c = 7.

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2 years ago

a hot tub pump circulate 4 cubic feet of water per minute, how many cubic inches of water does it circulate earch minute?

Shtirlitz [24]

$4 {ft }^{3} \times {( \frac{12in}{ft} )}^{3} = 6912 {in}^{3}$

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2 years ago

If x=5, evaluate x^5 * x^-3

Artyom0805 [142]

Answer: 25

Step-by-step explanation:

Plug in x to the expression.

$5^5 * 5^{-3}\\$

When there is the same base of exponents is being multiplied the exponents add together and the base stays the same.

So this simplifies to, $5^2$ .

This then solves to 25.

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3 years ago

A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra

Ksivusya [100]

The Quadrilateral is JKLM,

let $M_{JK}, M_{KL}, M_{LM}, M_{JM},$ be the midpoints of JK, KL, LM and JM respectively.

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Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

$M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})$ , </span>

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thus the coordinates of points $M_{JK}, M_{KL}, M_{LM}, M_{JM},$

are as follows:

$M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)$

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

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}$ </span>

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

thus the distances connecting the opposite entrances can be calculated as follows:

$|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5$

$|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39$

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m

8 0

3 years ago

Find the area of the following shape. You must show all work to receive credit

kirza4 [7]

Answer:

42 square units

Step-by-step explanation:

Draw segments - 5 to A, A to the point of intersection of Band C and from from the point of intersection of B, C upto 2. It will form a square with each side 8 units and four triangles.

Subtracting the area of 4 four triangles from the area of the square will give you the area of the required shape.

Area of the shape

$= {8}^{2} - \frac{1}{2} \times 3 \times 6 - \frac{1}{2} \times 5 \times 2 - \frac{1}{2} \times 2 \times 3\\ - \frac{1}{2} \times 2 \times 5 \\ \\ = 64 - 3 \times 3 - 5 \times 1 - 1\times 3 - 1 \times 5 \\ \\ = 64 - 9 - 5 - 3 - 5 \\ \\ = 64 - 22 \\ = 42 \: {units}^{2}$

3 0

3 years ago

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