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2 years ago

LORD PLEASE HELP MEEEEEEEEEEEEEEEEEEEEEE!

Mathematics

2 answers:

mars1129 [50]2 years ago

6 0

Answer:

( -1,7)

Step-by-step explanation:

{ refer attachment }

nekit [7.7K]2 years ago

5 0

Answer:

The answer would be (-1,7)

Step-by-step explanation:

Using the process of elimination we can subtract the two equations to cancel out the y variable. That is subtract the second equation from the first.

3x + y = 4

Now that we have found x we can substitute -1 for x into any equation for y. So, in substituting -1 for x in the second equation we get:

2(-1) + y = 5

-2 + y = 5

y = 5 + 2

y = 7

ANSWER: (-1, 7)

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Need help help on this question

balu736 [363]

Answer:70

Step-by-step explanation:

step 1:add up the angles you know

step 2:subtract the answer (220) from the total angle measure of a square which is 360 then you get 140

step 3: divide 140 by 2 because the two angles on the sides are equivalent and you get 70

6 0

3 years ago

10 points?...1) must be quick to answer,2) must be reasonable answer,3) Show explanation for your answer

Ilia_Sergeevich [38]

Your answer is
${x}^{2} \div {2} {z}^{9}$
because the numbers raised to negative powers must be flipped over the divisor to become positive. Then, when multiplying 2 and z, you add their exponents.

3 0

3 years ago

Given the position of the particle, what the position(s) of the particle when it’s at rest

choli [55]

The position function of a particle is given by:

$X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t$

The velocity function is the derivative of the position:

$\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}$

The particle will be at rest when the velocity is 0, thus we solve the equation:

$2t^2-9t-18=0$

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

$t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Substituting:

$\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}$

We have two possible answers: