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4 years ago

Find the area. Round to the nearest tenth.6 in.5 in.9 in.in.

Mathematics

1 answer:

ycow [4]4 years ago

7 0

Answer:

131 1/4 in.²

Step-by-step explanation:

To find the area of the figure, you would have to divide the figure into two parts. The figure can be divided into two rectangles.

<u>Rectangle 1</u>

The length is 11 3/4 in. The width is 6 in.

A = lw

A = (11 3/4 in.)(6 in.)

A = 70 1/2 in.²

<u>Rectangle 2</u>

The length is 9 in. The width is 6 3/4 in.

A = lw

A = (9 in.)(6 3/4 in.)

A = 60 3/4 in.²

Add the two areas together.

70 1/2 in.² + 60 3/4 in.² = 131 1/4 in.²

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2. Which statement about simplifying |-11 - 34 is TRUE?

zalisa [80]

Answer:

Step-by-step explanation:

-11-34 the answer will be -45 so option B is the best posible answer.

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1 year ago

Suppose y varies inversely with x. If y = 150 when x = 2, find y when x = 25.

IceJOKER [234]

Answer is 12

I hope this helps enough

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4 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

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3 years ago

ANSWER ASAP PLS 20 POINTS! Which graph represents the piecewise defined function? y= {3 if x<-2 0 if x=1 -1 if x>1

Otrada [13]

Answer:

It is the third option. (the one which has a black point at X=1, Y=1)

Step-by-step explanation:

For simplicity what you should do is, you should satisfy the second condition that is, y=0 if x=1.

Here only the third option is satisfying this condition.

Whenever there is is a white circle or a white point it means that the point does not satisfy the condition and the the function is not defined at that point. And the black point means that the circle is defined at that point and satisfies the condition.

Henceforth, the correct answer is the third option.

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4 years ago

* WILL GIVE BRAINLIEST* (06.04 LC)

sleet_krkn [62]

This is in slope-intercept form, y=mx+b. m is the slope (b is the intercept). So the slope here is 8.

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3 years ago

Find the area. Round to the nearest tenth.6 in.5 in.9 in.in.​

Find the area. Round to the nearest tenth.6 in.5 in.9 in.in.