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2 years ago

Use the coordinates J(7, 8), K(1, 2) and L(5, 2) for AJKL.

Mathematics

1 answer:

Lerok [7]2 years ago

6 0

Answer:

4.3

Step-by-step explanation:

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Pls pls help asap pls

Stolb23 [73]

Answer:

Step-by-step explanation:

3y+7=2x

3y=2x-7

y=(2/3)x-7/3

when parallel, the y=ax+b, the a keeps the same

so it‘s y=(2/3)x-a

and it pass (2,6), so 6=(2/3)*2 -a

6=4/3-a

-a=(18-4)/3

-a=14/3

a=-14/3

6 0

3 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

Need help ASAP, willing to give brainliest to the best answer!

sveta [45]

1. 70, 2. None, 3. Mike

6 0

3 years ago

Read 2 more answers

What is 648 divided by 4 in long division?

Rainbow [258]

The answer is 162

I hope this help

7 0

3 years ago

What is the volume of this pyramid? 945 cm³ 1260 cm³ 1890 cm³ 2520 cm³ A pyramid with a right triangular base. The right triangu

notsponge [240]

Answer:

The volume of the pyramid is $1,260\ cm^{3}$

Step-by-step explanation:

we know that

The volume of the triangular pyramid is equal to

$V=\frac{1}{3}BH$

where

B is the area of the triangular base

H is the height of the pyramid

step 1

Find the area of the base B

$B=\frac{1}{2}bh$

we have

$b=14\ cm$

$h=18\ cm$

substitute

$B=\frac{1}{2}(14)(18)=126\ cm^{2}$

step 2

Find the volume

we have

$B=126\ cm^{2}$

$H=30\ cm$

substitute

$V=\frac{1}{3}BH$

$V=\frac{1}{3}(126)(30)=1,260\ cm^{3}$

5 0

2 years ago

Read 2 more answers