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IgorLugansk [536]
3 years ago
8

PLEASEEE HELP MEEEE !!!!!

Mathematics
1 answer:
quester [9]3 years ago
5 0

Answer:

i-

Step-by-step explanation:

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You have just graduated from college and purchased a car for $8000. Your credit limit is $11,000. Assume that you make no paymen
Angelina_Jolie [31]
<span>assume the graduate put the purchase on his card.thenInitial balance = $8000 on the first statement (+fees and interest charges, if any)That means he owes the card issuer $8000.Credit balance is what the issuer owes the card holder, which is zero
</span> option "c. $8000" is your answer
8 0
4 years ago
Read 2 more answers
X+y=2 8x+3y=-19 substitution
padilas [110]

Answer:

{x,y}={−5,−7}

Explain:

// Solve equation [2] for the variable  y [2]    3y = 7x + 14 [2]    y = 7x/3 + 14/3 // Plug this in for variable  y  in equation [1] [1]    8x - 3•(7x/3+14/3) = -19 [1]    x = -5 // Solve equation [1] for the variable  x [1]    x = - 5 // By now we know this much : x = -5 y = 7x/3+14/3 // Use the  x  value to solve for  y y = (7/3)(-5)+14/3 = -7 Solution : {x,y} = {-5,-7}

Hoped I helped

4 0
3 years ago
Read 2 more answers
Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]

Answer:

Yes they are

Step-by-step explanation:

In the triangle JKL, the sides can be calculated as following:

  • J(2;5); K(1;1)

             => JK = \sqrt{(1-2)^{2} + (1-5)^{2}  } = \sqrt{(-1)^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • J(2;5); L(5;2)

             => JL = \sqrt{(5-2)^{2} + (2-5)^{2}  } = \sqrt{3^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • K(1;1); L(5;2)

             =>  KL = \sqrt{(5-1)^{2} + (2-1)^{2}  } = \sqrt{4^{2}+1^{2}  } = \sqrt{1+16}=\sqrt{17}

In the triangle QNP, the sides can be calculate as following:

  • Q(-4;4); N(-3;0)

             => QN = \sqrt{[-3-(-4)]^{2} + (0-4)^{2}  } = \sqrt{1^{2}+(-4)^{2}  } = \sqrt{1+16}=\sqrt{17}

  • Q (-4;4); P(-7;1)

   => QP = \sqrt{[-7-(-4)]^{2} + (1-4)^{2}  } = \sqrt{(-3)^{2}+(-3)^{2}  } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}

  • N(-3;0); P(-7;1)

             =>  NP = \sqrt{[-7-(-3)]^{2} + (1-0)^{2}  } = \sqrt{(-4)^{2}+1^{2}  } = \sqrt{16+1}=\sqrt{17}

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

7 0
3 years ago
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Someone answer this question I know your reading this.
IgorC [24]

Answer: A.

Step-by-step explanation:

To answer this, you must first understand what fractional exponents mean.

An exponent of 1 is simply the base itself. 7^1 = 7.

An exponent of 2 is squaring the base: 7^2 = 49.

An exponent of 1/2 is taking the square root of the base: 7^(1/2) = sq root of 7

Therefore, an exponent of 1/3 is taking the cube root of the base.

In this case, the base is 2.03 * t.

Therefore, this can be represented as the cube root of the product of 2.03 and the age of the mammal, so the answer is A.

3 0
3 years ago
What is the surface area of this square pyramid?
OLEGan [10]
240cm2 is the answer
3 0
3 years ago
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