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ivanzaharov [21]
3 years ago
15

Use the matrix tool to solve the system of equations. Choose the correct

Mathematics
1 answer:
posledela3 years ago
5 0
4x-9y=2 this is the answer
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Solve the proportion for x.
Delvig [45]

\frac{x}{8}  =  \frac{8}{64}  \\ 64x = 8 \times 8 \\ 64x = 64 \\ x =  \frac{64}{64}  \\  \boxed{x = 1}

  • Proportion for x is <u>1</u><u>.</u>
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5 0
2 years ago
White and black shapes are used in a game Some of the shapes are circles All the other shapes are squares The ratio of white to
nydimaria [60]
Is there a picture to this question??
5 0
4 years ago
Read 2 more answers
En un triangulo ABC, el angulo B mide 64° y el angulo C mide 72°. La bisectriz interior CD corta a la altura BH y a la bisectriz
nadezda [96]

Answer:

The difference between the greatest and the smallest angle of the triangle PBQ is 98°.

Step-by-step explanation:

The question is:

In a triangle ABC, angle B measures 64 ° and angle C measures 72°. The inner bisector CD intersects the height BH and the bisector BM at P and Q respectively. Find the difference between the greatest and the smallest angle of the triangle PBQ.

Solution:

Consider the triangle ABC.

The measure of angle A is:

angle A + angle B + angle C = 180°

angle A = 180° - angle B - angle C

             = 180° - 72° - 64°

             = 44°  

It is provided that CD and BM are bisectors.

That:

angle BCP = angle PCH = 36°

angle CBQ = angle QBD = 32°

angle BHC = 90°

Compute the measure of angle HBC as follows:

angle HBC = 180° - angle BHC + angle BCH

                  = 180° - 90° - 72°

                  = 18°

Compute the measure of angle BPC as follows:

angle BPC = 180° - angle PCB + angle CBP

                  = 180° - 18° - 36°

                  = 126°

Then the measure of angle BPQ will be:

angle BPQ = 180° - angle BPC

                  = 180° - 126°

angle BPQ = 54°

Compute the measure of angle PBQ as follows:

angle PBQ = angle B - angle QBD - angle HBC

                  = 64° - 32° - 18°

angle PBQ = 14°

Compute the measure of angle BQP as follows:

angle BQP = 180° - angle PBQ - angle BPQ

                  = 180° - 14° - 54°  

angle BQP = 112°

So, the greatest and the smallest angle of the triangle PBQ are:

angle BQP = 112°

angle PBQ = 14°

Compute the difference:

<em>d</em> = angle BQP - angle PBQ

  = 112° - 14°

  = 98°

Thus, the difference between the greatest and the smallest angle of the triangle PBQ is 98°.

4 0
3 years ago
What is the product of 4 2/3 and 11 1/4
mihalych1998 [28]

First you have to get rid of the mixed numbers, so multiply the whole numbers with the denominators and add the numerators so that way you get:

14/3 times 45/4....then cross simplify if possible and finally you can multiply the numerators and the denominators...

7/1 times 15/2 = 105/2....simplify it and get: 52 1/2


7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.
astra-53 [7]

Answer:

3\pi \rightarrow y=2\cos \dfrac{2x}{3}\\ \\\dfrac{2\pi }{3}\rightarrow y=6\sin 3x\\ \\\dfrac{\pi }{3}\rightarrow  y=-3\tan 3x\\ \\10\pi \rightarrow y=-\dfrac{2}{3}\sec \dfrac{x}{5}

Step-by-step explanation:

The period of the functions y=a\cos(bx+c) , y=a\sin(bx+c), y=a\sec (bx+c) or y=a\csc(bx+c) can be calculated as

T=\dfrac{2\pi}{b}

The period of the functions y=a\tan(bx+c) or y=a\cot(bx+c) can be calculated as

T=\dfrac{\pi}{b}

A. The period of the function y=-3\tan 3x is

T=\dfrac{\pi}{3}

B. The period of the function y=6\sin 3x is

T=\dfrac{2\pi}{3}

C. The period of the function y=-4\cot \dfrac{x}{4} is

T=\dfrac{\pi}{\frac{1}{4}}=4\pi

D. The period of the function y=2\cos \dfrac{2x}{3} is

T=\dfrac{2\pi}{\frac{2}{3}}=3\pi

E. The period of the function y=-\dfrac{2}{3}\sec \dfrac{x}{5} is

T=\dfrac{2\pi}{\frac{1}{5}}=10\pi

5 0
4 years ago
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