All categories

3 years ago

Please help me with this ! ! ! ! !

Mathematics

2 answers:

melisa1 [442]3 years ago

8 0

Answer:

First choice: m = -4

Step-by-step explanation:

The second equation is already solved for n, so lets's use the substitution method and substitute m - 2 for n in the first equation.

m - 2n = 8

m - 2(m - 2) = 8

m - 2m + 4 = 8

-m + 4 = 8

-m = 4

m = -4

kodGreya [7K]3 years ago

4 0

The second equation tells you what 'n' is. It's "m-2" .

So you can simply write "m-2" in place of 'n' in the first equation.

Then the first equation says [ m - 2(m-2) = 8 ], and you should have no trouble solving that.

You might be interested in

Please answer to receive ten points. I'll vote who answers first the brainliest if its correct.

beks73 [17]

The distance between points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ can be calculated using formula

$P_1P_2=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$

$AB=\sqrt{(-7-(-11))^2+(8-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\approx 5.657.$

$AC=\sqrt{(-4-(-11))^2+(4-4)^2}=\sqrt{49+0}=\sqrt{49}=7.$

$CB=\sqrt{(-7-(-4))^2+(8-4)^2}=\sqrt{9+16}=\sqrt{25}=5.$

Then

$P_{ABC}=5.657+7+5=17.657\approx 17.7$

Answer: correct choice is C.

3 0

2 years ago

Help againnn<br><br> i hate matg

ser-zykov [4K]

Hello User!

Correction "Math"

Answer: 12 4/7

6 0

2 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago

To input your angle measures in degrees, use the functions sind, cosd and tand, instead of sin, cos and tan. Zach stands at the

kramer

Answer:

about 2949 feet

Step-by-step explanation:

The geometry of the situation can be modeled by a right triangle. The height of the cliff can be taken to be the side opposite the given angle, and the distance to the coyote will be the side adjacent to the given angle. The relation between these values is the trig function ...

Tan = Opposite/Adjacent

<h3>setup</h3>

Filling in the known values, we have ...

tan(6°) = (310 ft)/(distance to coyote)

<h3>solution</h3>

Multiplying by (distance to coyote)/tan(6°) gives ...

distance to coyote = (310 ft)/tan(6°) ≈ 310/0.105104 ft

distance to coyote ≈ 2949.453 ft

The coyote is about 2949 feet from the base of the cliff.

6 0

2 years ago

A 32cm long string just fits round the perimeter of a square board. The area of the board is

Ierofanga [76]

Answer:

64cm2

Step-by-step explanation:

(32÷4)=8

8×8=64cm2

8 0

3 years ago

Read 2 more answers