40% of the students in a school are boys. lf there are 240 boys, how many student are therein the school?

The rectangle below is enlarged using a scale factor of 1.5. What will be the new length and width of the rectangle?

Likurg_2 [28]

Answer: length = 4.5 m, width = 3 m

Step-by-step explanation:

When a shape is enlarged using a scale factor, you simply multiply each of the sides by the scale factor.

length = 3 * 1.5 = 4.5 m

width = 2 * 1.5 = 3 m

5 0

3 years ago

giving the brainliest answer to who ever can help me out!!! AND has the correct answer need help asap!?!!

Afina-wow [57]

Answer:

caca water

pee water

Step-by-step explanation:

6 0

3 years ago

What is the slope of the line that passes through the points<br> (-16, -5) & (-15, -7)

In-s [12.5K]

Answer:

<u>slope=-2</u>

Step-by-step explanation:

Okay so we just take those coordinates and apply them to this formula:

y2-y1/x2-1

then simplify to get -2

if you want me to elaborate just say the word

8 0

2 years ago

The population in thousands of people of a city is approximated by the function P(t) = 1200(2)0.1038twhere t is the number of ye

asambeis [7]

The population of the group in 2019 is 2,134,000 people

Let P represent the population in thousands of people in the year t after 2011

Given the equation:

$P=1200(2)^{0.1038t}$

a) The population of the group in 2019, that is t = 8(2019 - 2011):

$P=1200(2)^{0.1038*8}=2134$

b) The population of the group in 2029, that is t = 18(2029 - 2011):

$P=1200(2)^{0.1038*18}=4382$

The population of the group in 2019 is 2,134,000 people

Find out more on equation at: brainly.com/question/2972832

7 0

2 years ago

Express the function as the sum of a power series by first using partial fractions. f(x) = 8 x2 − 4x − 12 f(x) = ∞ n = 0 find th

alexandr1967 [171]

I'm guessing the function is

$f(x)=\dfrac8{x^2-4x-12}=\dfrac8{(x-6)(x+2)}$

which, split into partial fractions, is equivalent to

$\dfrac1{x-6}-\dfrac1{x+2}$

Recall that for $|x|$ we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

With some rearranging, we find

$\dfrac1{x-6}=-\dfrac16\dfrac1{1-\frac x6}=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n$

valid for $\left|\dfrac x6\right|$ , or $|x|$ , and

$\dfrac1{x+2}=\dfrac12\dfrac1{1-\left(-\frac x2\right)}=\displaystyle\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

valid for $\left|-\dfrac x2\right|$ , or $|x|$ .

So we have

$f(x)=\displaystyle-\frac16\sum_{n=0}^\infty\left(\frac x6\right)^n-\frac12\sum_{n=0}^\infty\left(-\frac x2\right)^n$

$f(x)=\displaystyle-\sum_{n=0}^\infty\left(\frac{x^n}{6^{n+1}}+\frac{(-x)^n}{2^{n+1}}\right)$

$f(x)=\displaystyle-\sum_{n=0}^\infty\frac{1+3(-3)^n}{6^{n+1}}x^n$

Taken together, the power series for $f(x)$ can only converge for $|x|$ , or $-2$ .

6 0

3 years ago