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2 years ago

Factor to write an equivalent expression: 26a - 10

Mathematics

2 answers:

-BARSIC- [3]2 years ago

4 0

Answer:

2 (13a -5)

Step-by-step explanation:

26a - 10

Factor out a 2 from each term

2 (13a -5)

Burka [1]2 years ago

3 0

<h3>Answer: 2(13a-5)</h3>

Work Shown:

The GCF of 26 and 10 is 2. This is the largest factor that goes into both.

26a = 2*13a

10 = 2*5

So,

26a - 10 = 2*13a - 2*5 = 2(13a-5)

I'm using the distributive property to factor the 2 out.

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The perimeter of a rectangle is 13cm and its width is 11/4. find its length.

enyata [817]

Perimeter = 2 length + 2 width

P = 2(l) + 2(11/4)

$13 = 2l +2* \frac{11}{4}$

$13 = 2l + \frac{2}{1} * \frac{11}{4}$

$13 = 2l + \frac{22}{4}$ (subtract 22/4 from each side)

$13 -\frac{22}{4} = 2l$

$13 -\frac{11}{2} = 2l$

$\frac{26}{2} -\frac{11}{2} = 2l$

$\frac{15}{2}} = 2l$

$7.5 = 2l$ (divide each side by 2)

7.5 ÷ 2 = l

3.75 = l

The answer is 3.75 or $3 \frac{3}{4}$

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2 years ago

Evaluate 6+3b if b=7

Pavlova-9 [17]

27. Replace b with 7, making 6 + 3(7). The 3 and parentheses of 7 hint multiplication, so you multiply to make 21, then add 6 relating back to PEMDAS to teach you the correct order to solve the problem.

6 0

2 years ago

Read 2 more answers

A bakery sold 147 chocolate cupcakes in a day, which was 49% of the total number of cupcakes sold that day. How many total cupca

Mademuasel [1]

Answer:300

Step-by-step explanation:

7 0

1 year ago

the weight of a bag of golf balls varies directly as the number of golf balls in the bag.If a bag of 69 gold balls weighs 2,553

seropon [69]

Answer: 30

Step-by-step explanation:

Given :The weight of a bag of golf balls varies directly as the number of golf balls in the bag.

Let x be the number of golf balls in a bag that weighs 1,110 grams.

Then we have the following direct variation equation,

$\dfrac{x}{1110}=\dfrac{69}{2553}$

Multiply 1110 both sides , we get

$x=\dfrac{69}{2553}\times1110=30$

Hence, there are 30 balls in the bag.

4 0

1 year ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0

1 year ago