HELP PLEASE!!!!

Mathematics

1 answer:

brilliants [131]2 years ago

8 0

B. -1(7x+10)(x+2) expand and check

You might be interested in

Among all pairs of numbers with a sum of 238238, find the pair whose product is maximum. Write your answers as fractions reduced

mylen [45]

Answer:

$x=119119,\;\;\;y=119119$

Step-by-step explanation:

Let the $x$ be the first number, $y$ be the second number and $p$ be the product of the first and second number.

i.e $x+y=238238$ $\Rightarrow x=238238-y$

$p=xy$

$\Rightarrow p=(238238-y)y\\\Rightarrow p=238238y-y^2$

$\Rightarrow \frac{dp}{dy} =238238-2y\\\Rightarrow 0=238238-2y\\\Rightarrow 2y=238238\\\Rightarrow y=119119$

Now,

$x=238238-119119\\\\ \Rightarrow x=119119$

Hence, $x=119119,\;\;\;y=119119$

8 0

2 years ago

The endpoints of line segment RS are R(-2,-1), and S (2,3). Find the coordinates of the midpoint of line segment RS.

madam [21]

Midpoint coordinates are ( - 2 + 2) / 2 , (-1+3) / 2

= (0,1) Answer

6 0

3 years ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$