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3 years ago

5

A piece of pipe is 44 inches long. It is cut into three pieces. The longest piece is 3 times as long as the middle size piece, &

amp;’ the short is Piece is 26 inches shorter than the longest piece. Find the link of the three pieces of pipe

1 answer:

Dima020 [189]3 years ago

8 0

Sent a picture of the solution to the problem (s).

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Determine whether each expression can be used to find the length of side AB. Match Yes or No for each

tankabanditka [31]

Answer:

$(a)\ AB = \frac{7}{\sin (B)}$ $\to Yes$

$(b)\ AB = \frac{24}{\cos (B)}$ $\to Yes$

$(c)\ AB = \frac{24}{\cos (A)}$ $\to No$

$(d)\ AB = \frac{7}{\cos (A)}$ $\to Yes$

Step-by-step explanation:

Given

$BC =24$

$AC = 7$

Required

Select Yes or No for the given options

$(a)\ AB = \frac{7}{\sin (B)}$ $\to Yes$

Considering the sine of angle B, we have:

$\sin(B) = \frac{Opposite}{Hypotenuse}$

$\sin(B) = \frac{7}{AB}$

Make AB, the subject

$AB = \frac{7}{\sin(B)}$

$(b)\ AB = \frac{24}{\cos (B)}$ $\to Yes$

Considering the cosine of angle B, we have:

$\cos(B) = \frac{Adjacent}{Hypotenuse}$

$\cos(B) = \frac{24}{AB}$

Make AB the subject

$AB = \frac{24}{\cos(B)}$

$(c)\ AB = \frac{24}{\cos (A)}$ $\to No$

Considering the cosine of angle B, we have:

$\cos(A) = \frac{Adjacent}{Hypotenuse}$

$\cos(A) = \frac{7}{AB}$

Make AB the subject

$AB = \frac{7}{\cos(A)}$

$(d)\ AB = \frac{7}{\cos (A)}$ $\to Yes$

<em>This has been shown in (c) above</em>

3 0

3 years ago

Weekly demand for milk at a small grocery store is normally distributed with a mean of 251 gallons per week and standard deviati

slavikrds [6]

Solution

Given

mean = 251

standard deviation = 33

(a)

$\begin{gathered} P(Z$

Hence, X = 300

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2 years ago

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I need help w/ this!! thank you

Fittoniya [83]

<h3><u>Answer:</u></h3>

$\boxed{\pink{\sf \leadsto Yes \ there \ is \ a \ solution \ of \ the \ given \ inequality .}}$

<h3><u>Step-by-step explanation:</u></h3>

A inequality is given to us and we need to convert it into standard form and see whether if it has a solution . So let's solve the inequality.

The inequality given to us is :-

$\bf\implies |2y + 3 | - 1 \leq 0 \\\\\bf\implies |2y+3|\leq 1 \\\\\bf\implies (|2y+3|)^2 \leq 1^2 \\\\\bf\implies (2y+3)^2 \leq 1 \\\\\bf\implies (2y)^2+3^2+2(2y)(3) \leq 1 \\\\\bf\implies 4y^2+9+12y - 1 \leq 0 \\\\\bf\implies 4y^2+12y+8 \leq 0 \\\\\bf\implies 4( y^2 + 3y + 2 ) \leq 0 \\\\\bf\implies y^2+3y +2 \leq 0 \:\:\bigg\lgroup \purple{\bf Standard \ form \ of \ inequality }\bigg\rgroup \\\\\bf\implies y^2y+2y+y+2 \leq 0 \\\\\bf\implies y(y+2)+1(y+2)\leq 0 \\\\\bf\implies ( y+2)(y+1)\leq 0 \\\\\bf\implies \boxed{\red{\bf y \leq (-2) , (-1) }}$

Let's plot a graph to see its interval . Graph attached in attachment .

Now we can see that the Interval notation of would be ,

$\boxed{\boxed{\orange \tt \purple{\leadsto}y \in [-2,-1] }}$

<h3><u>Hence</u><u> the</u><u> </u><u>standa</u><u>rd</u><u> </u><u>form</u><u> </u><u>of</u><u> </u><u>inequa</u><u>lity</u><u> </u><u>is</u><u> </u><u>y²</u><u>+</u><u>3y</u><u> </u><u>+</u><u>2</u><u> </u><u>≤</u><u> </u><u>0</u><u> </u><u>and</u><u> </u><u>the </u><u>Solution</u><u> </u><u>set</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>ineq</u><u>uality</u><u> </u><u>is</u><u> </u><u>[</u><u> </u><u>-</u><u>2</u><u> </u><u>,</u><u> </u><u>-</u><u>1</u><u> </u><u>]</u><u> </u><u>.</u></h3>

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3 years ago

Simplify the expression<br> –2 + 7(x + 8)

anastassius [24]

Answer:

54+7x

Step-by-step explanation:

-2+7(x+8)

-2+7x+56

54+7x

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Pls help i dont get it

baherus [9]

Answer:

What is it?

Step-by-step explanation:

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