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3 years ago

Write an equation in slope intercept form for the line that passes through (1,8) and (2,11)

Mathematics

2 answers:

svet-max [94.6K]3 years ago

8 0

Answer:

both b and d cross those points

Step-by-step explanation:

solved them

SVETLANKA909090 [29]3 years ago

3 0

im not really sure but im tginking b

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Is 24/3 its simplest form.

Dmitry [639]

Answer:

Step-by-step explanation:

you can further divide it into 24/3 to get 8

8 0

3 years ago

Read 2 more answers

The sum of six fifths 6 5 and six timessix times a number is equal to four fifths 4 5 subtracted from seven timesseven times the

maw [93]

Answer :

The required number is 2.

Step-by-step explanation:

Given : The sum of six fifths and six times a number is equal to four fifths subtracted from seven times the number.

To find : The number ?

Solution :

Let the number be 'x'.

The sum of six fifths and six times a number i.e. $\frac{6}{5}+6x$

Four fifths subtracted from seven times the number i.e. $7x-\frac{4}{5}$

According to question,

$\frac{6}{5}+6x=7x-\frac{4}{5}$

$7x-6x=\frac{6}{5}+\frac{4}{5}$

$x=\frac{10}{5}$

$x=2$

The required number is 2.

5 0

3 years ago

Can someone help me

OleMash [197]

Answer:

1/2

Step-by-step explanation:

Because the solid triangle results in a smaller dashed triangle after the dilation, its scale factor will be a fraction. The length of the top side is 2 on the dashed triangle. It corresponds to the top side of the solid triangle which is 4.

2/4 = 1/2

The scale factor is 1/2 the size.

4 0

4 years ago

Help please I’ll mark as brainliest

lorasvet [3.4K]

Answer:

y= - 0.5x + -7

Step-by-step explanation:

............

4 0

3 years ago

2root3 sin^<a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="ac9eec">[email protected]</a> - <a href

Phantasy [73]

Answer:

Step-by-step explanation:

$2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3}) } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}$

$\alpha =2 n\pi+ \frac{\pi }{6} ,2n\pi +\frac{11\pi }{6} (general~solution)$

$or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )$

5 0

4 years ago