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3 years ago

Evaluate the expressions using the distributive property. ⅔(-7) + ⅔(-5)

Mathematics

1 answer:

Norma-Jean [14]3 years ago

7 0

-8
2/3(-21/3) + 2/3(-15/3)
-42/9 + -30/9
-72/9 = -8

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Over a period of 12 hours the temperature dropped 36°F what was the average change in temperature in one hour?

Alex17521 [72]

Answer:

3°F

Step-by-step explanation:

36 divided by 12 is 3

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3 years ago

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Please help!

Mrrafil [7]

Answer:

$\sf C. \quad \dfrac{1}{9}$

Step-by-step explanation:

Addition Law for Probability

$\sf P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Given:

$\sf P(A)=\dfrac{1}{3}=\dfrac{3}{9}$

$\sf P(B)=\dfrac{2}{9}$

$\sf P(A \cup B)=\dfrac{4}{9}$

Substitute the given values into the formula and solve for P(A ∩ B):

$\implies \sf P(A \cup B) = P(A)+P(B)-P(A \cap B)$

$\implies \sf \dfrac{4}{9} = \sf \dfrac{3}{9}+\dfrac{2}{9}-P(A \cap B)$

$\implies \sf P(A \cap B) = \sf \dfrac{3}{9}+\dfrac{2}{9}-\dfrac{4}{9}$

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$\implies \sf P(A \cap B) = \sf \dfrac{1}{9}$

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1 year ago

Read 2 more answers

In the graph below, Point A represents Owen's house, Point B represents David's house and Point C represents the school. Who liv

ycow [4]

Answer:

• David

• 4 miles

Explanation:

In the graph:

The given locations are:

• Owen's House, A(11,3)

• David's House, B(15,13)

• School, C(3,18)

We determine both Owen's and David's distance from the school using the distance formula.

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Owen's distance from school (AC)

$\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}$

David's distance from school (BC)

$\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}$

We see from the calculations that David lives closer to the school, and by 4 miles.

The graph below is attached for further understanding:

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3 years ago

What is the value of x? 3x^3 - 6 = 186

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3x to the third=180
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