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3 years ago

PLEASE HELP!

Mathematics

1 answer:

disa [49]3 years ago

3 0

Answer:

y=-5

Step-by-step explanation:

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Ratios equivalent to 4:12?

Arte-miy333 [17]

Ratios equivalent to 4:12 = 1:3

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4 years ago

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How do I convert standard form to point slope

Fittoniya [83]

The point-slope form:

$y-y_1=m(x-x_1)$

$(x_1,\ y_1)$ - given point

$m$ - given slope

The standard form:

$Ax+By=C$

$y-y_1=m(x-x_1)$ use distributive property

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$mx-y=mx_1-y_1$

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3 years ago

–4b – 5 = 3b + 9 I need help pls

Kitty [74]

The value of b in the equation is -2

<h3>What is an algebraic equation?</h3>

An algebraic equation is an expression that consists of numbers, variables with their arithmetic operations.

When solving this kind of algebraic equation, we move the like terms to a side and solve for the variable(unknown).

From the given algebraic equation:

-4b - 5 = 3b + 9

By collecting the like terms, we have:

-4b -3b = 5 + 9

-7b = 14

Divide both sides by (-7)

$\mathbf{\dfrac{-7b}{-7} = \dfrac{14}{-7}}$

x = -2

Learn more about algebraic equations here:

brainly.com/question/4344214

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2 years ago

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Solve by graphing u=v and 6u=2v-24

Vadim26 [7]

Answer:

$u=-6 \\ \\ v=-6$

Step-by-step explanation:

In this exercise, we have two equations, namely:

$u=v \ and \ 6u=2v-24$

And we are asked to solve this problem by graphing. In this way, we can write a system of linear equations in two variables, but first of all, let's rewrite:

$u=y \\ \\ v=x$

Then:

$\left\{ \begin{array}{c}y=x\\6y=2x-24\end{array}\right.$

So here we have two lines.

The first one is:

$\boxed{y=x}$

This line passes through the origin and has a slope $m=1$

The second one is:

$6y=2x-24 \\ \\ \therefore y=\frac{2x-24}{6} \\ \\ \therefore \boxed{y=\frac{1}{3}x-4}$

This line has a slope $m=\frac{1}{3}$ and cuts the y-axis at $b=-4$

By using graph tools, we get the graph shown below, then:

$x=-6 \\ \\ y=-6 \\ \\ \\ Since \ u=y \ and \ v=x, then: \\ \\ u=-6 \\ \\ v=-6$

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3 years ago

WHAT IS S=2PIR^2+2PIRH SOLVE FOR H

grigory [225]

$\\~\qquad\quad S\ =\ 2PIR^2+2PIRH\\\underline{-2PIR^2\qquad-2PIR^2\qquad\quad\ }\\\\ S-2PIR^2=2PIRH\\\\ 2PIRH\ =\ S-2PIR^2\\\underline{ :(2PIR)\qquad :(2PIR)}\\\\\\ H=\dfrac{S-2PIR^2}{2PIR}\\\\\\ H=\dfrac{S}{2PIR}-\dfrac{2PIR^2}{2PIR}\\\\\\ H=\dfrac{S}{2PIR}-R$

or:

$\\~\qquad\quad S\ =\ 2PIR^2+2PIRH\\\\~\qquad\quad S\ =\ 2PIR(R+H)\\\underline{:(2PIR)\qquad :(2PIR)}\\\\\\~\ \ \dfrac{S}{2PIR}\ =\ R+H\\\\\\ \dfrac{S}{2PIR}-R=H\\\\\\ H=\dfrac{S}{2PIR}-R$

4 0

4 years ago