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snow_tiger [21]
2 years ago
14

I need these answers please, I have a D this'll bring up my grade alot.

Mathematics
1 answer:
Nana76 [90]2 years ago
6 0

Answer: a=-1, -3

h=-3, 0

k=8, -4

Step-by-step explanation:

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Solve for b.<br> 7<br> b<br> 12<br> b= ✓ [?]<br> Pythagorean Theorem: a2 + b2 = c2<br> Enter
fomenos

Answer: b=\sqrt{95}

Step-by-step explanation:

To solve for b, we want to use the Pythagorean Theorem as given.

b and 7 are the legs, and 12 is the hypotenuse.

7^2+b^2=12^2      [exponent]

49+b^2=144     [subtract both sides by 49]

b^2=95               [square root both sides]

b=\sqrt{95}

Now we know b=\sqrt{95}.

6 0
2 years ago
If you drive 23 miles south, make a turn and drive 39 miles east, how far are you, in a straight line, from
Rom4ik [11]

Answer:

45.27 miles

Step-by-step explanation:

Pythagorean theorem

23^2 + 39^2 = d ^ 2

d = 45.27 miles

6 0
1 year ago
•** Plss help me **•
VashaNatasha [74]

Answer:

30

Step-by-step explanation:

You can use the triangle area theorem-

15x4(/2)=30

4 0
1 year ago
Someone please help me
bearhunter [10]
B. triangle and triangular prism
6 0
2 years ago
1)A System of equations is shown below . What is the solution to the system of equations? 5x+2y=-15 2x-2y=-6
meriva

Answer:

x= -3 and y= 0

Step-by-step explanation:

5x+2y=-15

<u>2x-2y=-6     </u>

<u>7x        =-21</u>

x= -3

Putting value of x in equation 1  

5(-3) +2y=-15

-15+2y= -15

2y= 0

y= 0

This can be solved with the help of matrices

In matrix form the above equations can be written in the form

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

Let

\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right] = A  \left[\begin{array}{ccc}x\\y\\\end{array}\right]  = X  and  \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]= B

Then AX= B

or X= A⁻¹ B

where  A⁻¹= adj A/ ║A║   where mod A≠ 0

adj A=  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]

║A║= ( 5*-2- 2*2)= -10-4= -14≠0

X= A⁻¹ B

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14  \left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]   \left[\begin{array}{ccc}-15\\-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]    =- 1/14     \left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]

 \left[\begin{array}{ccc}x\\y\\\end{array}\right]  =- 1/14 \left[\begin{array}{ccc}42\\0\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]

\left[\begin{array}{ccc}x\\y\\\end{array}\right]  = \left[\begin{array}{ccc}-3\\0\\\end{array}\right]

From here x= -3 and y= 0

Solution Set = [(-3,0)]

3 0
2 years ago
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