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boyakko [2]
3 years ago
7

Words on this photo plz awnser

Mathematics
1 answer:
alexandr402 [8]3 years ago
4 0
Gabriella  x
Her sister  x+6
Their mother  2*(x+6)=2x+12
Their aunt  (2x+12)+x=3x+12
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What is 9x+210=4x+140
Svetradugi [14.3K]

Answer:Final answer: (45,30)

Step-by-step explanation:

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4 years ago
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Find the zeros of the function.<br><br> f(x) = 9(x + 2)(x − 9.5)
Ber [7]

Answer:

x= -2, 9.5

Step-by-step explanation:

8 0
3 years ago
It costs $100 to manufacture and ship to a vendor a large batch of souvenir hats, but each hat that is in the shipment brings in
melamori03 [73]

Answer:

5x - 100

Step-by-step explanation:

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3 years ago
I really need help with this
natta225 [31]

Answer:

y=2x+7

Step-by-step explanation:

Equation of line: y = mx + b

where m is the slope, and b is the y-intercept

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4 years ago
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The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0
3 years ago
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