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3 years ago

PLLLLLLLLZZZZZZZ SOMEONE HELP ME

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

5 0

Answer: First, you start with the equation:

472

Since four students rode in cars, you want to eliminate this part of the problem to find how many students actually rode in the buses

So you would subtract

from both sides and get:

468

Finally, you would divide both sides by

and get

students on each bus

Step-by-step explanation: First, you start with the equation:

472

Since four students rode in cars, you want to eliminate this part of the problem to find how many students actually rode in the buses

So you would subtract

from both sides and get:

468

Finally, you would divide both sides by

and get

students on each bus

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What is the equations of the line passing through points (-3,7) and (9,-1)?

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y = -2/3x + 5 or A.

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7+(((3^2)*((-5)+1))/2

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3 years ago

A line that has a rise of -34 and a run of 20.

aleksandr82 [10.1K]

Answer:

$y=-1.7x-34$

Step-by-step explanation:

An equation of a line is given $y=mx+c$ , where m is the gradient or $\frac{Rise}{Run}$

The rise indicates the change in value of y (vertical change in the graph).

The run indicates the change in value of x (horizontal change in the graph).

Let point 1 of the graph be (x,y) = (0,-34) and point 2 be (x,y) = (20,0).

$m =\frac{Rise}{Run} \\ =\frac{-34}{20}\\ =-1.7$

Substitute point 1 into the graph to get the constant c.

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$c=-34$

∴ The equation of the line is, $y=-1.7x-34$

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3 years ago

Twenty-six students apply to serve as an usher at a school function. Eight of the those applying are freshmen, 7 are sophomores,

Gelneren [198K]

Answer:

Step-by-step explanation:

from usatestprep:

The situation is not an example of uniform probability because freshmen, sophomores, juniors, and seniors do not have equal probabilities of being selected.; Uniform probability → equal probability of being selected

P(freshman) =

; P(sophomore) =

7 /

; P(junior) =

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; P(senior) =

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3 years ago

Read 2 more answers

Karen notices that segment BC and segment EF are congruent in the image below:

LenaWriter [7]

Answer:

Givens:

$A(-2;2) \ B(-2;4) \ C(0;3) \ D(2;1) \ E(2;3) \ F(0;2)$
$BC \cong EF$
$\angle C \cong \angle F$

Using this information we could recur to SAS postulate to demonstrate the congruence of this pair of triangles. We already have one side and one angle congruent, we just need to demonstrate that $AC \cong DF$ , that will prove the complete congruence.

So, based on the given coordinates, we're able to calculate the length of each side and find if they are the same. We use this formula:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\\d_{AC} =\sqrt{(0-(-2))^{2}+(3-2)^{2}}=\sqrt{4+1}=\sqrt{5}\\\\\d_{DF} =\sqrt{(0-2)^{2}+(2-1)^{2}}=\sqrt{4+1}=\sqrt{5}$

Therefore, $AC \cong DF$ , because they have the same length.

Now, based on these congruences:

$BC \cong EF$
$\angle C \cong \angle F$
$AC \cong DF$

We demonstrate by the Side-Angle-Side (SAS) postulate that ΔABC ≅ ΔDEF.

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3 years ago