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3 years ago

Find the total area of the prism. 144 sq. in. 576 sq. in. 864 sq. in.

Mathematics

1 answer:

anastassius [24]3 years ago

7 0

Answer:

$864\ in^{2}$

Step-by-step explanation:

The figure is a cube

The surface area of a cube is equal to

$S=6b^{2}$

where

b is the length side of the cube

In this problem we have

$b=12\ in$

substitute

$S=6(12^{2})=864\ in^{2}$

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Nolan rides his bike 63 miles in 270 minutes. What is his average speed in miles per hour?

s2008m [1.1K]

63 miles in 270 minutes will mean that nolan rides his bike at a distance of 63/270 miles in one minute.
there are 60 minutes in an hour. So to find out the distance he covers in 1 hour, 63/270 × 60 will give you 14 miles.
Since the question is asking for the average speed, nolan rides his bike at 14 miles per hour i think.

4 0

3 years ago

Factorise x squared + 3x + 2

Zielflug [23.3K]

Answer:

$(x + 2)(x + 1)$

Step-by-step explanation:

by factorising ;

=》

${x}^{2} + 3x + 2$

=》

${x}^{2} + 2x + x + 2$

=》

$x(x + 2) + 1(x + 2)$

=》

$(x + 2)(x + 1)$

i hope it helped.....

8 0

2 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$