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3 years ago

Write the equation of the line in point-slope form that passes through (1, -4) and has a slope of 1/4

1 answer:

8 0

Hello!

Point-slope form is $y-b=m(x-a)$ , where m is the slope and (a, b) is a point on the graph.

You are given the slope ( $\frac{1}{4}$ ) and a point on the graph (1, -4), so you can just substitute the values into the form, like so:

$y-(-4)= \frac{1}{4} (x-1)$

Simplify:

$y+4= \frac{1}{4} (x-1)$

And that's it!

Answer:
$y+4= \frac{1}{4} (x-1)$

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Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as

stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

we have that

Jayden's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

Sheridan's Work

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

6 0

3 years ago

Read 2 more answers

Graph the equation y=-2x+1

Fofino [41]

Start by plotting the y-intercept at (0,1).

From that point, count "up 2, right 1" to get a second point on your graph.

If needed repeat that "up 2, right 1" from that second point to get a third point.

Draw the line that connects these 2 or 3 points.

6 0

3 years ago

Given 2x² = -8, x is ____

IrinaVladis [17]

Answer:

2x² = -8

x² = -4

The answer is no real solutions (x ∉ R) because a perfect square cannot be negative.

5 0

3 years ago

Read 2 more answers

Write an equation for the following: y varies directly with x and z. Find K when x=2, y=12 and z=5.

olga_2 [115]

$\qquad \qquad \textit{direct joint variation} \\\\ \textit{\underline{y} varies directly with \underline{x} and \underline{z}}~\hspace{8.5em} \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}xz~\hfill } \\\\ \textit{\underline{y} varies directly with \underline{x}, }w^2\textit{ and }z^5~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}xw^2z^5~\hfill } \\\\[-0.35em] ~\dotfill$

$y = kxz\qquad \textit{we know that} \begin{cases} x=2\\ y=12\\ z=5 \end{cases}\implies 12=k(2)(5) \\\\\\ \cfrac{12}{(2)(5)}=k\implies \cfrac{6}{5}=k~\hspace{10em}\boxed{y=\cfrac{6}{5}xz}$

6 0

2 years ago

I don't know the answer please help

KatRina [158]

I had this question today. I believe its either D or C, but my friend told me it was D. So, I think it's D. I hoped if this helped you or not. If you get it wrong Im very sorry so it would've been C.

5 0

3 years ago