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3 years ago

Part A: For each problem, write an expression using parenthesis, and then solve.

Mathematics

1 answer:

slavikrds [6]3 years ago

6 0

<span>1. $119 2. 64 beads 3. 4 candles 4</span>

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In a scale drawing, a building has a length of 15 cm. The actual length of the building is 37.5 feet. Whats the scale of the dra

Feliz [49]

Answer:

$\frac{15}{37.5}\ \frac{cm}{ft}$

$1.31:100$

Step-by-step explanation:

we know that

The scale of the drawing is equal to divide the length of a building in the scale drawing by the actual length of a building

$\frac{15}{37.5}\ \frac{cm}{ft}$

That means

15 cm in the drawing represent 37.5 ft in the actual

Remember that

$1\ ft=30.48\ cm$

Convert 15 cm to ft

divide by 30.48

$15\ cm=15/30.48\ ft$

substitute

$\frac{15}{37.5}\ \frac{cm}{ft}=\frac{(15/30.48)}{37.5}=\frac{0.0131}{1}$

$1.31:100$

That means

1.31 units in the drawing represent 100 units in the actual

5 0

3 years ago

Rewrite in Polar Form....<br> x^(2)+y^(2)-6y-8=0

skad [1K]

$\bf x^2+y^2-6y-8=0\qquad 
\begin{cases}
x^2+y^2=r^2\\\\
y=rsin(\theta)
\end{cases}\implies r^2-6rsin(\theta)=8
\\\\\\
\textit{now, we do some grouping}\implies [r^2-6rsin(\theta)]=8
\\\\\\\
[r^2-6rsin(\theta)+\boxed{?}^2]=8\impliedby 
\begin{array}{llll}
\textit{so we need a value there to make}\\
\textit{a perfect square trinomial}
\end{array}$

$\bf 6rsin(\theta)\iff 2\cdot r\cdot \boxed{3\cdot sin(\theta)}\impliedby \textit{so there}
\\\\\\
\textit{now, bear in mind we're just borrowing from zero, 0}
\\\\
\textit{so if we add, \underline{whatever}, we also have to subtract \underline{whatever}}$

$\bf [r^2-6rsin(\theta)\underline{+[3sin(\theta)]^2}]\quad \underline{-[3sin(\theta)]^2}=8\\\\
\left. \qquad \right.\uparrow \\
\textit{so-called "completing the square"}
\\\\\\\
[r-3sin(\theta)]^2=8+[3sin(\theta)]^2\implies [r-3sin(\theta)]^2=8+9sin^2(\theta)
\\\\\\
r-3sin(\theta)=\sqrt{8+9sin^2(\theta)}\implies r=\sqrt{8+9sin^2(\theta)}+3sin(\theta)$

3 0

3 years ago

Is it a, b, c, or d?

soldier1979 [14.2K]

Answer:

The answer is B. 4,7.5,8

Step-by-step explanation:

Becouse A and C Both of them is Right-angled triangle and D is Obtuse triangle

7 0

3 years ago

As a first step in solving the system shown, Yumiko multiplies both sides of the equation 2x – 3y = 12 by 6. By what factor shou

GenaCL600 [577]

Answer:

Yumiko should multiply the other equation by 3.

If she adds the two equations she would be left with the variable 'x'.

Step-by-step explanation:

Given the two equations are as follows:

$2x - 3y = 12 \hspace{5mm} \hdots (1)$

$5x + 6y = 18 \hspace{5mm} \hdots (2)$

It is given that she multiplies the first equation by 6. Therefore, (1) becomes

$12x - 18y = 72 \hspace{15mm} \hdots (a)$

Now, note that the sign of the variable 'y' is negative. So, if we make the co-effecient of 'y' equal in both the cases, add them it would result in the elimination of the variable 'y'.

The co-effecient of y in Equation (2) is 6. To make it 18 like it is in Equation (1), we multiply throughout by 3.

Therefore, Equation (2) becomes:

$15x + 18y = 54 \hspace{5mm} \hdots (b)$