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3 years ago

6

Denise has $365 in her saving account. She wants to save at least $635. Write and solve an inequality to determine how much more

money Denise must save to reach her goal. Let d represent the amount of money in dollars Denise must save to reach her goal.

1 answer:

Charra [1.4K]3 years ago

6 0

she must save 370 dollars

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My favorite Pokemon card has a mass of about one and a half

s344n2d4d5 [400]

Answer:

grams cause its is really small

7 0

3 years ago

In a basketball game Jane and Francine made 20 baskets. Jane scored 4 times as many as Francine. How many baskets did Jane score

VMariaS [17]

Alright, lets get started.

Suppose Francine made number of baskets = x

As given in question, Jane scored 4 times as many as Francine.

Means Jane has scored number of baskets will be = 4x

Jane and Francine made 20 baskets together, means

$x + 4x = 20$

$5x = 20$

Dividing 5 in both sides

$\frac{5x}{5} = \frac{20}{5}$

x = 4

Means Francine made number of baskets = 4

Hence Jane made number of baskets = $4 * 4 = 16$ : Answer

Hope it will help :)

3 0

3 years ago

Read 2 more answers

7 1/4 + 5 3/5= reduced to it's lowest form

Scrat [10]

7 1/4 + 5 3/5
First, you need a common denominator to add the two. 20 will work.
4x5=20, so we have to also multiply 1 by 5.
4x5=20, 1x5=5
5x4=20, so we have to also multiply 3 by 4.
5x4=20, 3x4=12
And now we put the fractions back in.
7 5/20 + 5 12/20
And add them.
7 5/20 + 5 12/20 = 12 17/20
Now, we're supposed to reduce it to its lowest form.
Unfortunately, the fraction cannot be reduced, because 17 is a prime number, meaning there are no factors to it except one and itself.
Therefore, the lowest form of 12 17/20 is 12 17/20
Your answer is B, 12 17/20

4 0

4 years ago

PLS HELP I WILL GIVE BRAINLIEST

Goryan [66]

Answer:

I think it is the second one i am so so so sorry if it wrong

Step-by-step explanation:

8 0

3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent <em>p</em>-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

(c) The series

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago