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2 years ago

Twice the difference between a number and 4 is twenty. Find the number.

Mathematics

2 answers:

kogti [31]2 years ago

7 0

Answer:

D 14

Step-by-step explanation:

Let x be the number

2(x-4) = 20

Divide each side by 2

2/2(x-4) = 20/2

x-4 = 10

Add 4 to each side

x-4+4 = 10+4

x = 14

Vikki [24]2 years ago

5 0

Answer:

The number is 14

Step-by-step explanation:

Let's call this unknown number of x:

2×(x – 4) = 20

2x – 8 = 20

2x = 20 + 8

2x = 28

x = 28/2

x = 14

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Georgia [21]

Answer:

y = 10m

Step-by-step explanation:

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y = centimeters

y = millimeters * 10

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3 years ago

In the rectangle ABDF above, C and E are midpoints of sides line ED and line DF, respectively, what fraction of the areas of the

aivan3 [116]

Answer:

Is shaded 5/8 of the area of the rectangle

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The area of triangle ABC represent 1/4 of the total area

The area of triangle CDE represent 1/8 of the total area

The shaded region is composed by one triangle equal to triangle ABC and three triangles equal to triangle CDE

The area of the shaded region is equal to

$(1/4)(1)+(1/8)(3)=(1/4)+(3/8)=(2/8)+(3/8)=5/8$

therefore

Is shaded 5/8 of the area of the rectangle

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3 years ago

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kati45 [8]

Answer:

Step-by-step explanation:

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2 years ago

Find the exact value of each trigonometric function for the given angle θ.

Kay [80]

Answer:

$\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.$

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since $\theta=240$ , it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

$\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}$

$\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}$

$\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}$

$\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}$

$\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2$

$\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}$

Therefore, $\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.$

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