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4 years ago

One fifth the sum of one half and one third

2 answers:

8 0

Rewrite the equation as
1/5 * (1/2 + 1/3)
We must first solve the stuff inside the parentheses, to do this we must find a common denominator, in this case 6

1/2 becomes 3/6 and 1/3 becomes 2/6 thus (1/2 + 1/3) = (3/6 + 2/6) = (5/6)
Now we can multiple (1/5)*(5/6) you can cancel out the 5's and then you are left with you answer
1/6

Juliette [100K]4 years ago

3 0

1/5( 1/2 + 1/3)
1/5 (3/6 + 2/6)
1/5(5/6)
5/30
1/6

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Please help been stuck on this hw question ALLL DAYYYYYY

Rama09 [41]

Answer:

1 5/9

Step-by-step explanation:

9 1/3 can be rewritten as 28/3. The question is asking for you to divide 28/3 by 6. 6 is 6/1, so the expression is 28/3 divided by 6/1. You then do the reciprocal of 6/1 and the expression is $\frac{28}{3} * \frac{1}{6}$ . This gets you 28/18, which is simplified to 14/9 or 1 5/9. Hope this helps!

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3 years ago

Read 2 more answers

Billy has a gallon of paint. He us going to pour it into a paint tray that measures 10 inches wide, 14 inches long and 4cm deep.

erica [24]

Answer:

Tray will overflow with $10.53\ \text{inch}^3$ of paint.

Step-by-step explanation:

Dimensions the tray is 10 inch by 14 inch by 4 cm

$1\ \text{cm}=\dfrac{1}{2.54}\ \text{inch}$

$4\ \text{cm}=\dfrac{4}{2.54}\ \text{inch}$

Volume the tray can hold is

$10\times 14\times \dfrac{4}{2.54}\ \text{in}^3$

$1\ \text{inch}^3=\dfrac{1}{231}\ \text{gallon}$

The volume of paint Billy has is $1\ \text{gallon}$

Difference in the volume of paint and volume of tray in cubic inches is

$\left(1-\dfrac{10\times14\times\dfrac{4}{2.54}}{231}\right)231=10.53\ \text{inch}^3$

The tray will overflow with $10.53\ \text{inch}^3$ of paint.

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3 years ago

Please help I need P factored completely <br> P(x)=x^4+50x^2+625

victus00 [196]

Answer: here u go and can u give me brainliest

Step-by-step explanation:

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3 years ago

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cricket20 [7]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Here's the solution :

9. Statement : $\tt{VB\:\: bisects\;\; \angle EVO}$

Reason : $\tt Given$

10. Statement : $\tt \angle3 \cong \angle 4$

Reason : $\tt Definition \;\; of \;\; angle \;\; bisector$

11. Statement $\tt{VB\:\: bisects\;\; \angle EBO}$

Reason : $\tt Given$

12. Statement : $\tt \angle1 \cong \angle 2$

Reason : $\tt Definition \;\; of \;\; angle \;\; bisector$

13. Statement : $\tt \overline{BV }\cong \overline{BV}$

Reason : $\tt Common\:\: side$

14. Statement : $\tt \triangle \:BEV \cong \triangle \:BOV$

Reason : $\tt ASA \:\: postulate$

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7 0

2 years ago

PLEASE HELP 100 POINTS!!!!!!

horrorfan [7]

Answer:

A) See attached for graph.

B) (-3, 0) (0, 0) (18, 0)

C) (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

$g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad \textsf{if }x\geq 3\end{cases}$

Therefore, the function has two definitions:

$g(x)=x^3-9x \quad \textsf{when x is less than 3}$

$g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}$

When <u>graphing</u> piecewise functions:

Use an open circle where the value of x is <u>not included</u> in the interval.
Use a closed circle where the value of x is <u>included</u> in the interval.
Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=(3)^3-9(3)=0 \implies (3,0)$

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)$

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

$\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}$

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

$\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}$

$\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b$

$\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}$

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation: (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

3 0

2 years ago