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3 years ago

How do you solve for y?

Mathematics

1 answer:

ss7ja [257]3 years ago

4 0

I’m pretty sure you just solve exactly like that
-7(-2)-1 =13 at least I’m told that \_( •-• )_/

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Question 2

BabaBlast [244]

The correct answer is choice d because you would divide x by 2 to get 70. then you do the same with y so 35/2 = 17.5

7 0

2 years ago

50 points decreased by 26%

labwork [276]

26% of 50
26/100 * 50
13

50 - 13 = 37

7 0

3 years ago

Read 2 more answers

What is the place value for 3 in 318,472,008

VikaD [51]

Salutations!

What is the place value for 318,472,008

The number 3 is in the place as hundred million.

In the numeric form, we can write the number as ------

300, 000, 000.

Hope I helped.

6 0

3 years ago

the product of two integers is positive.is the quotient of those same integers positive, negative,or impossible to predict? expl

KiRa [710]

Answer:

Step-by-step explanation:

pi is irrational. That was proven long ago. Unfortunately the proof is harder to understand than the proof that the square root of 2 is irrational, which is accessible to most high school students.

To prove that p is irrational you need to understand calculus, up through integration. Even then, the proof is rather intense and takes careful attention.

8 0

2 years ago

Match each set of vertices with the type of triangle they form.

Andrew [12]

Answer: The calculations are done below.

Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

$AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.$

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

$AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.$

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

$AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.$

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

$AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.$

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

$AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.$

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

4 0

3 years ago

Read 2 more answers