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2 years ago

Beth is half as old as Meg. Enter an algebraic expression based on this situation.

Mathematics

2 answers:

Tanya [424]2 years ago

8 0

Answer:

b = 1/2m

Step-by-step explanation:

Let:

b = beth's age
m = meg's age

The prompt tells us that Beth is half as old as Meg. Therefore, mathematically, b = 1/2m.

lara31 [8.8K]2 years ago

6 0

The answer is b=1/2 m. I know the question was already answered but I need to answer more questions to ask questions lol.

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Math question 3, thanks if you help:)

Zanzabum

Answer:

Step-by-step explanation:

Plug in x = 2

g(2) = 3(2)² - 2*2 + 5

= 3* 4 - 4 + 5

= 12 - 4 + 5

= 8 + 5

= 13

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3 years ago

Find the solution set for the following absolute value equation:<br> [3x + 9] = 24

zalisa [80]

Answer:

x = -11 or 5

Step-by-step explanation:

[3(-11)+9]

[-33+9]

[-24]

[3(5)+9]

[15+9]

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3 years ago

Can you please answer the whole page I’ll give you all the points I have and Brainiest answer

MAVERICK [17]

QUESTION 1

The given circle has equation:

$x^2+y^2=16$

We can rewrite to obtain;

$(x-0)^2+(y-0)^2=4^2$

This is equation is now in the form;

$(x-h)^2+(y-k)^2=r^2$

Where (h,k)=(0,0) is the center of the circle and $r=4$ is the radius.

QUESTION 2

The center of the circle is at (1,1) and passes through the point (4,5).

The radius of this circle is given by;

$r=\sqrt{(4-1)^2+(5-1)^2}$

$r=\sqrt{(3)^2+(4)^2}$

$r=\sqrt{9+16}$

$r=\sqrt{25}$

$r=5$ units.

We substitute into the formula;

$(x-h)^2+(y-k)^2=r^2$

$(x-1)^2+(y-1)^2=5^2$

$(x-1)^2+(y-1)^2=25$

Expand if you wish;

$x^2-2x+1+y^2-2y+1=25$

$x^2+y^2-2x-2y+2-25=0$

$x^2+y^2-2x-2y-23=0$

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3 years ago

Rhe quotient of the sum of 2 and a number b, and 3

Mamont248 [21]

"The quotient of the sum of 2 and number b, and 3" can be displayed as:

(2 + b)/3

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3 years ago

Is this correct? The answer I picked was the only one that had no limit. If I did it correctly the other choices limit was zero.

andriy [413]

Step-by-step explanation:

These are all examples of p-series:

∑(1 / nᵖ), where p>0.

If p > 1, the series converges. If 0 < p ≤ 1, the series diverges.

First option:

∑(1/n⁵)

Here, p = 5. Since 5 > 1, the series converges.

Second option:

∑((√n+3)/n³)

∑((√n)/n³) + ∑(3/n³)

∑(1/n^2.5) + 3 ∑(1/n³)

In the first sum, p = 2.5. In the second sum, p = 3. Both are greater than 1, so the series converges.

Third option:

∑((n−4)/(n⁴√n))

∑(1/(n³√n)) − ∑(4/(n⁴√n))

∑(1/n^3.5) − 4 ∑(1/n^4.5)

In the first sum, p = 3.5. In the second sum, p = 4.5. Both are greater than 1, so the series converges.

Fourth option:

∑(1/∛n)

∑(1/n^⅓)

Here, p = ⅓. This is less than 1, so the series diverges.

Note: if a series is converging, then the limit is 0.

However, if the limit of a series is 0, it does not necessarily mean that series is converging.

Here, the limit of all 4 options is 0. However, the fourth option is a diverging series.

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4 years ago