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2 years ago

8

Wilma used 60 centimeters of lace to make a border around a rectangular card. The width of the card is 20 centimeters. What is t

he length of the card?

1 answer:

iris [78.8K]2 years ago

4 0

60 - (20 + 20)= 20
20

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The altitude of a plane is 1200 meters. From a point on the ground, the angle of elevation of an airplane is 25º. What is the di

gtnhenbr [62]

Answer:

2839.4 meters

Step-by-step explanation:

Given that:

Altitude = 1200 m

Using trigonometry :

The distance from point P to the airplane :

Using trigonometric relation :

Sin θ = opposite / hypotenus

Sin θ = altitude / x

Sin θ = 1200 m / x

Sin 25 = 1200 / x

0.4226182 = 1200 / x

x = 1200 / 0.4226182

x = 2839.4423

Distance from P to airplane = 2839.4 meters

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2 years ago

Which of the following compositions will map the red figure onto the blue figure?<br> 6lB<br> ATA

melisa1 [442]

Answer:

Point A is translated to A" by T(0, -6), since A has been shifted 6 units down

check the picture, the translated triangle is A"'B"'C"'.

clearly the angle between B"'A"' and B"A"' is 90 degrees, so we can complete the transformation by a 90° clockwise rotation or a 270° counterclockwise rotation.

from the choices given, we see that is is a 270° counterclockwise rotation.

Answer: A.(o,-6). Ra,270

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Step-by-step explanation:

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2 years ago

F(x) = –22 + 10x - 6<br> Find f(-5)

Darina [25.2K]

Answer:

Step-by-step explanation:

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2 years ago

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Identify the maximum and minimum values of the function y = 10 cos x in the interval [-2π, 2π]

grigory [225]

Hello,

-1<=cos x <=1
==>-10<=10*cos x <=10
max y=10 if x=0 +2kπ ==>x=0,-2π,2π
min y=-10 if x=π+2kπ ==>x=-π,π

6 0

2 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

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

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

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

so

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

so

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

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

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

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

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

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

step 4

Find sin(θ+β)

we know that

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

so

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

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

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

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

2 years ago