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3 years ago

Hey yall, 20 points? Factor to write an equivalent expression: 26a−10

Mathematics

1 answer:

Oksanka [162]3 years ago

7 0

Answer:

2 (13a -5)

Step-by-step explanation:

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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emmasim [6.3K]

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4 years ago

1.) 3(x + 7) - 9x (x = 1

MrMuchimi

Answer:

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3 years ago

On a coordinate plane, the x-axis is labeled bags of trail mix and the y-axis is labeled ounces of almonds. Line a is labeled y

Musya8 [376]

Answer:

Line B

Step-by-step explanation:

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3 years ago

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Help me plzzzz!!!!<br><br> Graph: y-3=1/2(x+2)

Zolol [24]

Answer:

Step-by-step explanation:

First you need to find the Slope Intercept form or find the point and slope.

The point is: (-2,3)

and the slope is: 1/2

y - 3 = 1/2(x + 2)

y - 3 = 1/2x + 1

y = 1/2x + 1 + 3

y = 1/2x + 4

Slope intercept form: y = 1/2x + 4

Graph is show below ---

8 0

3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago