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3 years ago

Solve for xO 5.5O 6.5O 7.5O 8.5

Mathematics

1 answer:

Leni [432]3 years ago

4 0

Answer:

x= 120

Step-by-step explanation:

x+60=180

x=180-60

x=120

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How much is $1250 worth at the end of 3 year, if the interest rate of 6.5% is compounded weekly?

matrenka [14]

Answer:

3 X 52 =156

100+6.5 = 106.5, 106.5/100 = 1.065

1250 X 1.065^156 = 23091323.5

$23091323.50

6 0

3 years ago

2: Write out a system of equations that has infinite solutions. Explain why it has infinite

Rasek [7]

Answer:

Below!

Explanation:

A system of equations with infinite solutions defines that both the equations are identical and are overlapping when the lines are graphed. An example could be y = 5x + 9 and y = 5x + 9. These sets of equations have infinite solutions because they are the same and when graphed, they overlap.

Hoped this helped!

4 0

2 years ago

5(3m+2) 6(g+h) 4d+8 21p+35q 18x+9y

aalyn [17]

(a) 5(3m+2) = 15m + 10

(b) 6(g+h) = 6g + 6h

(c) 4d+8 = 4(d + 2)

(d) 21p+35q = 7(3p + 5q)

(e) 18x+9y = 9(2x + y)

7 0

3 years ago

si entre 2 obreros tardan 10 días en hacer una carretera, ¿cuánto tardarán en hacer el mismo trabajo 4 obreros?

Olegator [25]

Answer:

5 dias

Step-by-step explanation:

10/2= 5

8 0

3 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

Solve for xO 5.5O 6.5O 7.5O 8.5​

Solve for xO 5.5O 6.5O 7.5O 8.5