Please help me solve this

Mathematics

2 answers:

Kobotan [32]3 years ago

7 0

The answer is : y-21 = -21 / 11 (x-7)
hope that helped !

amm18123 years ago

6 0

The answer is :

Y-21=-21/11 (X -7)

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Find the value of d.

Ket [755]

First find the value or (DEA)
For this you will have to use the Pythagorean theory,
${a}^{2} + {b}^{2} = {c}^{2}$
C is mainly the hypotenuse,
Meaning you either want to find A or B,

Since the hypotenuse has two segments, both values are 10, then just add them together making it 20,

Then 16 will be A or B, in this case it will be A,

To find a you must first move A to the other side by doing the opposite,
${b}^{2} = {c}^{2} - {a}^{2}$
Now plug in the values,
${b}^{2} = {20}^{2} - {16}^{2}$
${b}^{2} = 400 - 256$
${b}^{2} = 144$
Now just find the square root of both factors,
$\sqrt{ {b}^{2} } = \sqrt{12}$
The value of (DEA) is,
$b = 12$
Now we want to find (d)
Since 12 is the value of the base, then divide it 2 to find (DE), after just repeat the whole process but with the value of hypotenuse 10 instead of 20 since we want to find the smaller triangle,
${a}^{2} = {c}^{2} - {b}^{2}$
Now plug in the value
${a}^{2} = {10}^{2} - {6}^{2}$
${a}^{2} = 100 - 36$
${a}^{2} = 64$
Now find the square root of both factors,
$\sqrt{ {a}^{2} } = \sqrt{64}$
$a = 8$

The value of (d) is (8)

Hope this helped
:D

4 0

3 years ago

How many 4 card hands that contain exactly 3 aces and 1 queen can be chosen from a 52 card deck?

sesenic [268]

There are 16 possible hands.

Choosing 3 aces from 4 possible is expressed by:
$_4C_3=\frac{4!}{3!1!} = 4$

Choosing 1 queen from 4 possible is expressed by:
$_4C_1=\frac{4!}{1!3!}=4$

This gives us 4*4 = 16 possible hands.

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3 years ago

Lilit [14]

We first find A2, for which we multiply A with itself. We perform matrix multiplication in which we multiply rows of first matrix with columns of second matrix element by element and add.

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3 years ago

PLZZ HELP<br> What is the solution to: y=-2x+7 and y=3x+2?

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Please see the pic, I'd solved in it.