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Orlov [11]
3 years ago
11

Question 4 (1 point)

Mathematics
1 answer:
igomit [66]3 years ago
4 0

Answer:

x = 15

Step-by-step explanation:

We want to solve for x in  3*(2x + 5) = 3x + 4x

First simplify.

3*(2x + 5) = 7x

Next, distribute the 3.

3*2x + 3*5 = 7x

6x + 15 = 7x

15 = 7x - 6x

15 = x

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Liono4ka [1.6K]
33 is the answer :) Please give me the brainliest answer, and a rate and thanks.
5 0
3 years ago
One afternoon in January, the temperature outside was -4*F. By midnight, the temperature had dropped by 10*F. what was the tempe
loris [4]
The temperature at midnight would be -14*F. Subtract 10 from -4 and you get -14*F
3 0
2 years ago
√₂
snow_lady [41]

Answer:

$2000 was invested at 5% and $5000 was invested at 8%.

Step-by-step explanation:

Assuming the interest is simple interest.

<u>Simple Interest Formula</u>

I = Prt

where:

  • I = interest earned.
  • P = principal invested.
  • r = interest rate (in decimal form).
  • t = time (in years).

Given:

  • Total P = $7000
  • P₁ = principal invested at 5%
  • P₂ = principal invested at 8%
  • Total interest = $500
  • r₁ = 5% = 0.05
  • r₂ = 8% = 0.08
  • t = 1 year

Create two equations from the given information:

\textsf{Equation 1}: \quad \sf P_1+P_2=7000

\textsf{Equation 2}: \quad \sf P_1r_1t+P_2r_2t=I\implies 0.05P_1+0.08P_2=500

Rewrite Equation 1 to make P₁ the subject:

\implies \sf P_1=7000-P_2

Substitute this into Equation 2 and solve for P₂:

\implies \sf 0.05(7000-P_2)+0.08P_2=500

\implies \sf 350-0.05P_2+0.08P_2=500

\implies \sf 0.03P_2=150

\implies \sf P_2=\dfrac{150}{0.03}

\implies \sf P_2=5000

Substitute the found value of P₂ into Equation 1 and solve for P₁:

\implies \sf P_1+5000=7000

\implies \sf P_1=7000-5000

\implies \sf P_1 = 2000

$2000 was invested at 5% and $5000 was invested at 8%.

Learn more about simple interest here:

brainly.com/question/27743947

brainly.com/question/28350785

5 0
1 year ago
0.09×0.7 <br>help me please
dem82 [27]

0.09x0.7= 0.063

That its the answer

7 0
3 years ago
The height h of a projectile is a function of the time t it is in the air. the height in feet for t seconds is given by the func
alexgriva [62]

Domain means the values of independent variable(input) which will give defined output to the function.

Given:

The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function

h(t)=-16t^2 + 96t

Solution:

To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.

To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq  0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq  0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible  \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)

Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq  0\\-112 \geq  0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq  0\\80 \geq  0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq  0\\-112 \geq  0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution

Conclusion:

The domain of the function is the time in between 0 to 6 seconds

0 \leq  t \leq  6

The height will be positive in the above interval.

7 0
3 years ago
Read 2 more answers
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