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expeople1 [14]
3 years ago
6

Please HELP ME WITH THIS TY!

Mathematics
1 answer:
LenKa [72]3 years ago
7 0
X=117 degrees
Y = 63 degrees
Z = 117 degrees
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I need answer quickly
Anton [14]

Answer:

√85

Step-by-step explanation:

a² + b² = c²

6² + b² = 11²

36 + b² = 121

b² = 121 - 36

b² = 85

b = √85

4 0
2 years ago
Read 2 more answers
4x^2-x-3=0<br> Solve by completing the square.<br> x= ?
rodikova [14]
<span>4x^2-x-3=0
let us rewrite the equation in much simplified form:
x^2 - 0.25x - 0.75 = 0
</span><span>x^2 - 0.25x = 0.75
</span>x^2 - 1/4 x + 1/64 =3/4 + 1/64
(x-1/8)^2 = 49/64
x−<span>1/8</span>=±<span>49/64

</span><span><span>x1</span>=−<span>3/4
</span></span><span><span>x2</span>=1</span>
6 0
3 years ago
[ALGEBRA 2 POLYNOMIAL DIVISION/REMAINDER THEOREM]
castortr0y [4]

Answer:

  k = -4

Step-by-step explanation:

The remainder theorem tells you that the remainder from division f(x)/(x -k) is f(k). You want the value of k such that ...

  f(k) = -15

Looking on the given graph, you find that k must be -4. That is, ...

  f(-4) = -15

  k = -4

_____

The divisor with k=-4 is (x -(-4)) = (x +4). The second attachment shows the division of f(x) by (x+4). The remainder is shown on the bottom line.

7 0
3 years ago
If Jeff sold 39 pumpkins, how much money did he make?
katrin2010 [14]
$195 each pumpkin cost about $5 each
8 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

The scale factor is s=\dfrac{1}{3} and the center of dilation is at the point (4,2).

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

According to the given information, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let us assume the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

5 0
3 years ago
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