Zoom in more, but can someone please help me with this?

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

6 0

The answer is A, because graph A shows her saving $3.00 per week up until she has the amount she needs. Hope this helped!

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Help guys make p da subject of the formula

Basile [38]

Answer:

$p=\dfrac{m}{k-s}$

Step-by-step explanation:

When they ask you to make a variable the "subject of the formula", they're basically asking you to build a "machine" for it. In the expression $y=mx+b$ , for instance, if we know what m, x, and b are, we can plug them all in and out comes a value for y! To build that machine, we need to get it by itself on one side of the equation.

So, we want to build this machine for p, and the equation we have to build it out of is

$1=\dfrac{pk}{m+sp}$

One think we can immediately say about this equation: for a fraction to be equal to one, its numerator and denominator have to be equal to each other. So we can say

$m+sp=pk$

Ok, what next? Well, if we want p by itself, we need to somehow peel it off of that sp and pk. Let's collect them on the right side by subtracting sp from both sides:

$m=pk-sp$

p is a common factor of both sp and pk, so we can peel it away using the distributive property:

$m=p(k-s)$

And from here, we can divide both sides by $k-s$ and we'll have p as our subject:

$\dfrac{m}{k-s}=p$ or $p=\dfrac{m}{k-s}$

So if we know m, k, and s, we can easily solve for p. If m = 4, k = 2, and s = 0:

$p=\frac{4}{2-0} =\frac{4}{2}=2$

3 0

2 years ago

Given: <SPT = <RQT, ST = RT Prove: PR = QS

andrey2020 [161]

Answer:

See explanation

Step-by-step explanation:

Consider triangles PTS and QTR. In these triangles,

$ST=RT$ - given;
$\angle SPT=\angle RQT$ - given;
$\angle STP=\angle RTQ$ - as vertical angles when lines PR and SQ intersect.

Thus, $\triangle PTS\cong \triangle QTR$ by AAS postulate.

Congruent triangles have congruent corresponding sides, so

$PT=QT$

Consider segments PR and QS:

$PR=PT+TR\ [\text{Segment addition postulate}]\\ \\QS=QT+TS\ [\text{Segment addition postulate}]\\ \\PT=QT\ [\text{Proven}]\\ \\ST=RT\ [\text{Given}]$