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3 years ago

12

Tom mowed lawns for the summer. In June he deposited $120 the first week, $80 the second week, and $75 the following week. Then

the mower needed repairs, so Tom had to withdraw $30 from his account. How much money did he have left in his account at the end of June?

1 answer:

melisa1 [442]3 years ago

4 0

So, let's add up the total amount he made:
120 + 80 + 75 = 275
If he withdrew $30:
275 - 30 = 245
Tom will have $245 at the end of June

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Write an equation in standard form of the hyperbola described.

marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

$\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill$

$\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}$

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2 years ago

Two passenger train, A and B, 450 km apart, star to move toward each other at the same time and meet after 2hours.

likoan [24]

Let $v$ be the speed of train A, and let's set the origin in the initial position of train A. The equations of motion are

$\begin{cases}s_A(t) = vt\\s_B(t) = -\dfrac{8}{7}vt+450\end{cases}$

where $s_A,\ s_B$ are the positions of trains A and B respectively, and t is the time in hours.

The two trains meet if and only if $s_A=s_B$ , and we know that this happens after two hours, i.e. at $t=2$

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3 years ago

3. There are 124 acorns laying on the ground in your yard, it is predicted that the number of acorns will

Gennadij [26K]

Answer:

507,409

Step-by-step explanation:

if every 5 days the number quadruples (x4), and we want to know how many acorns fall after 30 days, we can divide 30 by 5 so we only have to calculate for the amount of time the take to quadruple.

30 ÷ 5 = 6

so we only have to quadruple the acorns 6 times.

if we start off with 124 acorns, this is what it will look like:

Day 0: 124 acorns

Day 5: 124 x 4 = 496 acorns

Day 10: 496 x 4 = 1984 acorns

Day 15: 1984 x 4 = 7936 acorns

etc... until day 30.

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I hope this was helpful :-)

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3 years ago

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Dahasolnce [82]

Step-by-step explanation:

2x-12x+36 =90

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3 years ago

Some types of algae have the potential to cause damage to river ecosystems. Suppose the accompanying data on algae colony densit

Phantasy [73]

Answer:

$y=-2.95836 x +234.56159$

Step-by-step explanation:

We assume that th data is this one:

x: 50, 55, 50, 79, 44, 37, 70, 45, 49

y: 152, 48, 22, 35, 43, 171, 13, 185, 25

a) Compute the equation of the least-squares regression line. (Round your numerical values to five decimal places.)For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i =50+ 55+ 50+ 79+ 44+ 37+ 70+ 45+ 49=479$

$\sum_{i=1}^n y_i =152+ 48+ 22+ 35+ 43+ 171+ 13+ 185+ 25=694$

$\sum_{i=1}^n x^2_i =50^2 + 55^2 + 50^2 + 79^2 + 44^2 + 37^2 + 70^2 + 45^2 + 49^2=26897$

$\sum_{i=1}^n y^2_i =152^2 + 48^2 + 22^2 + 35^2 + 43^2 + 171^2 + 13^2 + 185^2 + 25^2=93226$

$\sum_{i=1}^n x_i y_i =50*152+ 55*48+ 50*22+ 79*35+ 44*43+ 37*171+ 70*13+ 45*185+ 49*25=32784$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=26897-\frac{479^2}{9}=1403.556$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=32784-\frac{479*694}{9}=-4152.22$

And the slope would be:

$m=-\frac{-4152.222}{1403.556}=-2.95836$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{479}{9}=53.222$

$\bar y= \frac{\sum y_i}{n}=\frac{694}{9}=77.111$

And we can find the intercept using this:

$b=\bar y -m \bar x=77.1111111-(-2.95836*53.22222222)=234.56159$

So the line would be given by:

$y=-2.95836 x +234.56159$

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2 years ago