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3 years ago

Which other parts must be equal for triangle ABC= triangle MTR by

Mathematics

1 answer:

lesya692 [45]3 years ago

7 0

Answer:

see below

Step-by-step explanation:

ASA stands for Angle-side-Angle

thus, for ∆ABC $\cong$ ∆MTR

$\displaystyle \angle C \cong \angle R$

SAS stands for Side-angle-Side 

thus,for ∆ABC $\cong$ ∆MTR

$\displaystyle \overline{ MR }\cong \overline{MT}$

likewise,

AAS stands for Angle-Angle-Side

thus,for ∆ABC $\cong$ ∆MTR

$\displaystyle \angle M \cong \angle T$

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Find the equation of the line that passes through the points (2, 8) and (-2, 6).

OLEGan [10]

The correct answer is C. y = 1/2x + 7

In order to find this, we first have to look for the slope. We can do that by using the slope formula below.

m(slope) = (y2 - y1)/(x2 - x1)

m = (8 - 6)/(2 - -2)

m = 2/4

m = 1/2

Now we can solve using the slope intercept form and a given point.

y = mx + b

8 = 1/2(2) + b

8 = 1 + b

7 = b

Now we can put these into the equation and model as y = 1/2x + 7

4 0

4 years ago

Rachel has 60 scones. She sells them to Robbie, Cameron, Louis, Tom and

Nutka1998 [239]

We can start by writing all that we know about the problem. Let us write down how many scones each boy has:
$R_{o} : x$
$C_{a} : x+k$
$L_{o} : x+2k$
$T_{o} : x+3k$
$C_{h} : x+4k$
Note that k is the increase, we know it's a constant. This is not a system of equations, please keep that in mind. We will use rest of the information given to build the system. We have two variables x (the number of cookies the first boy bought) and k( the increase), that means we need to have two equations to solve this problem.
We know that total amount of scones is 60, so we can use that information to build our first equation. We sum all of the scones and we get 60.
$R_{o}+C_{a}+L_{o}+T_{o}+C_{h}=60$
$5x+10k=60$
We know that Robbie and Cameron combined have 3/7 of Louis, Tom and Charlie combined. We can use this information to build our second equation.
 $R_o+C_a= \frac{3}{7} (L_o+T_o+C_h)$
$2x+k= \frac{3}{7} (x+2k+x+3k+x+4k)$
$2x+k= \frac{3}{7} (3x+9k)$
$14x+7k=9x+27k$
$5x=20k$
$x=4k$
Now this piece of information alows us to solve the first equation that we made and this will solve our problem. We simply plug in x=4k into 5x+10k=60.
$5(4k)+10k=60$
$30k=60$
$k=2$
Now we can plug this information into the same equation 5x+10k=60 and get x.
$5x+10(2)=60$
$5x=40$
$x=8$
And that's it. Now we can write out how many cookies each boy has bought.
$R_{o} : x=8$
$C_{a} : x+k=10$
$L_{o} : x+2k=12$
$T_{o} : x+3k=14$
$C_{h} : x+4k=16$
If you add them all up you get 60.

3 0

3 years ago

HURRY ILL GIVE BRAINLIEST!!!

Free_Kalibri [48]

ANSWER

D) (a + 1, b + m)

EXPLANATION

The given line passes through (a,b) and has slope m.

The equation is:

$y - b = m(x - a)$

When x=a+1, we obtain

$y - b = m(a + 1 - a)$

This

$y = m( 1 ) + b$

$y = m + b$

Therefore,(a + 1, b + m) must also lie on this line.

The correct answer is D.

3 0

3 years ago

Now, I know that real numbers are composed of _____,_____,_______,______, and ________

Over [174]

Answer:

real numbers are composed of

natural numbers, whole numbers, integers, rational numbers and irrational numbers.

Step-by-step explanation:

hope that helps>3

7 0

3 years ago

Read 2 more answers

Sat scores. sat scores of students at an ivy league college are distributed with a standard deviation of 250 points. two statist

erik [133]

The confidence interval for confidence level of $1-\alpha$ is

$\left(\overline x-Z_{\alpha/2}\dfrac\sigma{\sqrt n},\overline x+Z_{\alpha/2}\dfrac\sigma{\sqrt n}\right)$

where $\overline x$ is the sample mean, $Z_{\alpha/2}$ is the critical value for the given confidence level, $\sigma$ is the standard deviation of the population, and $n$ is the sample size. The margin of error is the $Z_{\alpha/2}\dfrac\sigma{\sqrt n}$ term.

a) For a confidence level of $1-\alpha=0.90$ , we have $Z_{\alpha/2}=Z_{0.05}\approx1.64$ . So in order to have a margin of error of at most 25 points, we have

$1.64\dfrac{250}{\sqrt n}=25\implies n\approx268.96$

so Raina should collect a sample of at least 269 students.

b) A confidence interval with a higher confidence level would more closely approximate and reflect the population, so it stands to reason that Luke should collect a larger sample than Raina to meet his 99% confidence spec.

c) For a confidence level of $1-\alpha=0.99$ , we have $Z_{\alpha/2}=Z_{0.005}\approx2.58$ . Then the margin of error would at most satisfy

$2.58\dfrac{250}{\sqrt n}=25\implies n\approx665.64$

so that Luke should collect a sample of at least 666 students.

4 0

3 years ago