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3 years ago

Anyone can help me ?

Mathematics

1 answer:

qaws [65]3 years ago

5 0

9514 1404 393

Answer:

∠x = 80°
∠y = 40°
∠z = 60°

Step-by-step explanation:

The angle marked x and the one marked 100° form a linear pair, so are supplementary. The measure of x is ...

x = 180° -100° = 80°

The angle marked 40° and the one marked y are "alternate interior angles" with respect to the parallel lines and the diagonal that crosses them. As such, they have the same measure:

y = 40°

The angle marked 100° and the sum of angles in the lower right corner are "corresponding" angles, so have the same measure. That is ...

100° = 40° +z

z = 100° -40° = 60°

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Nezavi [6.7K]

I think it's 12 but not really sure only because I never really learned this

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3 years ago

Solve √5r - 9 - 3 = √r +4-2

Mekhanik [1.2K]

Answer:

r = 5

Step-by-step explanation:

Solving radical equations often results in extraneous solutions. These can be avoided by using a graphing calculator for the solution.

<h3>Graph</h3>

The attached shows that the value of r that makes both sides of the equation have the same value is r = 5.

<em>Check:</em>

√(5·5 -9) -3 = √(5 +4) -2 ⇒ 4 -3 = 3 -2 . . . true

<h3>Analytical solution</h3>

Solution of an equation like this is typically done by isolating the radical expressions, then squaring the equation.

We can add 3, square the equation, then isolate the radical and square again:

$\displaysyle \sqrt{5r-9}-3=\sqrt{r+4}-2\\\\\sqrt{5r-9}=\sqrt{r+4}+1\qquad\text{add 3}\\\\5r-9=(r+4)+2\sqrt{r+4}+1\qquad\text{square both sides}\\\\(4r-14)^2=4(r+4)\qquad\text{subtract $(r+5)$ and square again}\\\\16r^2-116r+180=0\qquad\text{rewrite in standard form}\\\\4(4r -9)(r -5) = 0\qquad\text{factor}\\\\r=\{\dfrac{9}{4},\ 5\}\qquad\text{solutions to the factored form}$

Trying these values in the original equation, we get ...

√((5(9/4) -9) -3 = √(9/4 +4) -2 ⇒ 1.5 -3 = 2.5 -2 . . . . . false

√(5(5) -9) -3 = √(5 +4) -2 ⇒ 4 -3 = 3 -2 . . . . . true

The solution is r = 5.

8 0

2 years ago

5 feet<br> 4 feet<br> What is the perimeter?<br> 16 feet<br> 14 feet<br> 18 feet

brilliants [131]

perimiter means all the way around, and im assuming this is a 4-sided shape, so 5+5+4+4 = 18 ft.

8 0

3 years ago

If a 150 pound woman would receive 24 grams of a certain medicine. How many grams should a 120 pound woman receive?

kumpel [21]

With a problem like this, set up a proportion (like when making lemonade from a mix).

$\frac{weight}{medicine} = \frac{weight}{medicine}$ is our proportion.

We fill in the the two weights and the 24 grams of medicine in the proportion and solve for the fourth piece not there.

$\frac{150}{24} =\frac{120}{M}$

We shorten medicine to M and we solve for M by cross multiplying.

$150M = 2880$

$M = \frac{2880}{150} = 19.2 grams$

The woman should get 19.2 grams of medicine.

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4 years ago

If A and B are two angles in standard position in Quadrant I, find cos( A +B ) for the given function values. sin A = 8/17 and c

horsena [70]

Answer:

Part 1) $cos(A + B) = \frac{140}{221}$

Part 2) $cos(A - B) = \frac{153}{185}$

Part 3) $cos(A - B) = \frac{84}{85}$

Part 4) $cos(A + B) = -\frac{36}{85}$

Part 5) $cos(A - B) = \frac{63}{65}$

Part 6) $cos(A+ B) = -\frac{57}{185}$

Step-by-step explanation:

<u><em>the complete answer in the attached document</em></u>

Part 1) we have

$sin(A)=\frac{8}{17}$

$cos(B)=\frac{12}{13}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{8}{17})^2=1$

$cos^2(A)+\frac{64}{289}=1$

$cos^2(A)=1-\frac{64}{289}$

$cos^2(A)=\frac{225}{289}$

$cos(A)=\pm\frac{15}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{15}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{13})^2=1$

$sin^2(B)+\frac{144}{169}=1$

$sin^2(B)=1-\frac{144}{169}$

$sin^2(B)=\frac{25}{169}$

$sin(B)=\pm\frac{25}{169}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{5}{13}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{15}{17} \frac{12}{13}-\frac{8}{17}\frac{5}{13}$

$cos(A + B) = \frac{180}{221}-\frac{40}{221}$

$cos(A + B) = \frac{140}{221}$

Part 2) we have

$sin(A)=\frac{3}{5}$

$cos(B)=\frac{12}{37}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{3}{5})^2=1$

$cos^2(A)+\frac{9}{25}=1$

$cos^2(A)=1-\frac{9}{25}$

$cos^2(A)=\frac{16}{25}$

$cos(A)=\pm\frac{4}{5}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{4}{5}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{12}{37})^2=1$

$sin^2(B)+\frac{144}{1,369}=1$

$sin^2(B)=1-\frac{144}{1,369}$

$sin^2(B)=\frac{1,225}{1,369}$

$sin(B)=\pm\frac{35}{37}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{35}{37}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{4}{5} \frac{12}{37}+\frac{3}{5} \frac{35}{37}$

$cos(A - B) = \frac{48}{185}+\frac{105}{185}$

$cos(A - B) = \frac{153}{185}$

Part 3) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A-B)

we know that

$cos(A - B) = cos(A) cos(B)+sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A-B)

substitute in the formula

$cos(A - B) = \frac{8}{17} \frac{3}{5}+\frac{15}{17} \frac{4}{5}$

$cos(A - B) = \frac{24}{85}+\frac{60}{85}$

$cos(A - B) = \frac{84}{85}$

Part 4) we have

$sin(A)=\frac{15}{17}$

$cos(B)=\frac{3}{5}$

Determine cos (A+B)

we know that

$cos(A + B) = cos(A) cos(B)-sin(A) sin(B)$

step 1

Find the value of cos(A)

Remember that

$cos^2(A)+sin^2(A)=1$

substitute the given value

$cos^2(A)+(\frac{15}{17})^2=1$

$cos^2(A)+\frac{225}{289}=1$

$cos^2(A)=1-\frac{225}{289}$

$cos^2(A)=\frac{64}{289}$

$cos(A)=\pm\frac{8}{17}$

The angle A belong to the I quadrant, the cosine is positive

$cos(A)=\frac{8}{17}$

step 2

Find the value of sin(B)

Remember that

$cos^2(B)+sin^2(B)=1$

substitute the given value

$sin^2(B)+(\frac{3}{5})^2=1$

$sin^2(B)+\frac{9}{25}=1$

$sin^2(B)=1-\frac{9}{25}$

$sin^2(B)=\frac{16}{25}$

$sin(B)=\pm\frac{4}{5}$

The angle B belong to the I quadrant, the sine is positive

$sin(B)=\frac{4}{5}$

step 3

Find cos(A+B)

substitute in the formula

$cos(A + B) = \frac{8}{17} \frac{3}{5}-\frac{15}{17} \frac{4}{5}$

$cos(A + B) = \frac{24}{85}-\frac{60}{85}$

$cos(A + B) = -\frac{36}{85}$

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4 0

4 years ago