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3 years ago

Help 5 ratee????????????

Mathematics

1 answer:

masha68 [24]3 years ago

5 0

The measure of the other a he is 55 degrees

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Mrs. Smith is creating a rectangular flower bed such that the width is half of the

Brilliant_brown [7]

Answer:

As for this problem, we will first establish that the length of the flower bed be represented as x, the width of the flower bed be represented as x/2 ,and the area of the flower bed be taken as it is since it is given. We then follow the formula for area which is length multiplied to width which is:

A = LW

we then substitute them

34 square feet = x (x/2)

now all we need to do is find x first.

34 square feet = x squared / 2

now do a cross multiplication

68 square feet = x squared

then get the square root of both sides

8.246 feet = x

Since x is equal to the length of the flower bed, all we have to do to get the width of it is to divide it by 2. So...

W = x/2

W = 8.246 feet / 2

W = 4.123 feet

And since the problem asked it to find the width of the flower bed to the nearest tenth of a foot, the answer would be 4.1 ft.

6 0

3 years ago

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

Alexei wants to hang a mirror in his boat, and put a frame around it. The mirror and frame must have an area of 19.25 square fee

Drupady [299]

<span>The first part is already in squared terms. "19.25 sq ft" Taking the measurements for the mirror 3ft and 5ft, you'd get: "15 sq ft." Subtract that answer from the previous number to get "4.25 sq ft.

Hope this helps out a little lol :)</span>

5 0

3 years ago

the length of a flag is 0.3 foot less than twice its width. if the perimeter is 14.4 feet longer than the width, find the dimens

forsale [732]

$x-\ length\\
y-\ width\\\\
x+0.3=2y\ \ \ \ \ | subtract\ 0.3\\x=2y-0.3\\
perimeter=2x+2y\\ 2x+2y-14.4=y\\\\
2*(2y-0.3)+2y-14.4=y\\
4y-0.6+2y-14.4=y\\
6y-15=y\ \ \ | add\ 15\\
6y=15+y\ \ | subtract\ y\\
5y=15\ \ \ | divide\ by\ 5\\
y=3\\
x=2y-0.3=2*3-0.3=6-0.3=5.7\\\\
Dimensions:\ length=5.7\ feet\, width=3\ feet.$