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3 years ago

Reflection. Can anyone figure this out?

Mathematics

1 answer:

wlad13 [49]3 years ago

5 0

Hope this helps :p

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John can ride his skate board 100 feet in 4.6 seconds. how fast is this in miles per hour?

diamong [38]

About .24 mile per hour
$60 \div 4.6 = 13.04 \\ 13.04 \times 100 = 1304.34$
1 mile is 5280 foot

4 0

3 years ago

To show that polygon ABCDE is congruent to polygon FGHIJ, a must be used to make the two polygons coincide. A sequence of two tr

Sati [7]

Hello.

The minimum number of rigid transformations required to show that polygon ABCDE is congruent to polygon FGHIJ is 2 (translation and rotation).

A rotation translation must be used to make the two polygons coincide.

A sequence of transformations of polygon ABCDE such that ABCDE does not coincide with polygon FGHIJ is a translation 2 units down and a 90° counterclockwise rotation about point D

Have a nice day

7 0

3 years ago

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A square garden has a side length of 11 meters. What is the area of the garden? What is the perimeter?

Talja [164]

Answer:

Area of the garden=121 square meter

perimeter of the garden=44 meter

Step-by-step explanation:

Area of a square=side*side

=11*11

=121square meter

Perimeter of a square garden=side*4

=11*4

=44 m

6 0

3 years ago

Read 2 more answers

8(3x+1) please give me your Brainly answer

kodGreya [7K]

Answer:

its 3

Step-by-step explanation:

Simplifying

8 = 3x + -1

Reorder the terms:

8 = -1 + 3x

Solving

8 = -1 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

8 + -3x = -1 + 3x + -3x

Combine like terms: 3x + -3x = 0

8 + -3x = -1 + 0

8 + -3x = -1

Add '-8' to each side of the equation.

8 + -8 + -3x = -1 + -8

Combine like terms: 8 + -8 = 0

0 + -3x = -1 + -8

-3x = -1 + -8

Combine like terms: -1 + -8 = -9

-3x = -9

Divide each side by '-3'.

x = 3

Simplifying

x = 3 this took forever to type hope it helped

5 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