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algol13
3 years ago
11

Could you help me quick!?

Mathematics
1 answer:
anastassius [24]3 years ago
4 0

Answer:

1: 5                       2: -70.98  3: C   4:A.     5:A

Step-by-step explanation:

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Please hurry i need this fast
damaskus [11]

Answer:

B!!! 9!!!

Step-by-step explanation:

5 0
3 years ago
Which statements are true about the ordered pair (10, 5) and the system of equations?
Alexeev081 [22]

Answer:

C

Step-by-step explanation:

We have the system of equations:

\left\{\begin{array}{ll}2x-5y=-5 \\ x+2y=11\end{array}

And an ordered pair (10, 5).

In order for an ordered pair to satisfy any system of equations, the ordered pair must satisfy both equations.

So, we can eliminate choices A and B. Satisfying only one of the equations does not satisfy the system of equations.

Let’s test the ordered pair. Substituting the values into the first equation, we acquire:

2(10)-5(5)\stackrel{?}{=}-5

Evaluate:

20-25\stackrel{?}{=}-5

Evaluate:

-5\stackrel{\checkmark}{=} -5

So, our ordered pair satisfies the first equation.

Now, we must test it for the second equation. Substituting gives:

(10)+2(5)\stackrel{?}{=} 11

Evaluate:

20\neq 11

So, the ordered pair does not satisfy the second equation.

Since it does not satisfy both of the equations, the ordered pair is not a solution to the system because it makes at least one of the equations false.

Therefore, our answer is C.

3 0
3 years ago
I need help with this !!!!!
GenaCL600 [577]

Answer:

-48

Step-by-step explanation:

-6(8)=-48

5 0
3 years ago
Read 2 more answers
Mathematics Mh Slope
gizmo_the_mogwai [7]

Answer:y=-3x+4

Step-by-step explanation: use the y=mx+b equation.

6 0
3 years ago
Read 2 more answers
Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required
xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

                A = Pe^{rt}

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               6000 = 3000e^{0.085t}

               \frac{6000}{3000} = e^{0.085t}

               2 = e^{0.085t}

Take log_{e} of both sides

               log_{e}2 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln2 = 0.085t

               t = \frac{ln2}{0.085}

               t = \frac{0.693}{0.085}

               t = 8.15

(b) For the amount to double, A = 3 × P

               A = 3 × $3000

               A = $9000

               9000 = 3000e^{0.085t}

               \frac{9000}{3000} = e^{0.085t}

               3 = e^{0.085t}

Take log_{e} of both sides

               log_{e}3 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln3 = 0.085t

               t = \frac{ln3}{0.085}

               t = \frac{1.0986}{0.085}

               t = 12.92

5 0
3 years ago
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