All categories

3 years ago

The first one is X, THE SECOND ONE IS Y, AND THE LAST ONE IS Z

Mathematics

1 answer:

DanielleElmas [232]3 years ago

6 0

The shape is nonexistent if there are two 60 degree angles the last angle must be 60 degrees and not 40

You might be interested in

The bike shop is having a sale. Everything in the store is 20% off. A bike is originally priced at $345.00 What is the sale of t

zubka84 [21]

Answer:

Sale price = $276

Step-by-step explanation:

Sale price = original price - discount

Discount = original price * percent off

Discount = 345*.2 = 69

The discount is 69 dollars

Sale price = 345- 69

= 276

5 0

3 years ago

Simplify -8(4-3x)-17x

Luda [366]

1. distribute the -8 into the parentheses and you get -32+24x-17x
2. combine like terms and get -32+7x
3. if you need to make it simpler set the equation = to zero -32+7x=0 and then add the 32 on both sides
4. 7x=32 and then divide by 7 and you get 32/7 or approximately 4.571

I hope this helped!! :)

4 0

3 years ago

Read 2 more answers

In this problem, y = c1ex + c2eâx is a two-parameter family of solutions of the second-order DE y'' â y = 0. Find a solution of

SCORPION-xisa [38]

Answer:

$y(t) = 2e^t -e^{-t}$

Step-by-step explanation:

Assuming this complete problem: "In this problem,

y = c1ex + c2e−x

is a two-parameter family of solutions of the second-order DE

y'' − y = 0.

Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

y(0) = 1, y'(0)= 3"

Solution to the problem

For this case we have a homogenous, linear differential equation with order 2, and with the general form:

$ay'' +by' +cy=0$

Where $a =1, b=0, c=-1$

And we can rewrite the differential equation in terms $y = e^{rt}$ like this:

$[e^{rt}]'' -e^{rt}=0$

And applying the second derivate we got:

$r^2 e^{rt} -e^{rt}=0$

We can take common factor $e^{rt}$ and we got:

$e^{rt} (r^2-1) =0$

And for this case the two only possibel solutions are $r=1, r=-1$

And the general solution for this case is given by:

$y = c_1 e^{r_1 t} + c_2 e^{r_2 t}$

Replacing the roots that we found we got:

$y = c_1 e^{t} +c_2 e^{-t}$

Now we can find the derivates for this last espression

$y' = c_1 e^t -c_2 e^{-t}$

$y'' = c_1 e^t +c_2 e^{-t}$

From the initial conditions we have this:

$y(0)=1 =c_1 e^{0} +c_2 e^{-0}= c_1 +c_2$ (1)

$y'(0) =3= c_1 e^{0} -c_2e^{-0}= c_1 -c_2$ (2)

If we add equations (1) and (2) we got:

$4 = 2c_1 , c_1 = 2$

And solving for $c_2$ we got:

$c_2=3-c_1= 3-2 = 1$

So then our general solution is given by:

$y(t) = 2e^t -e^{-t}$

8 0

3 years ago

A fashion store is having a sale advertising 'one third' off all prices. If a pair of trousers costs £54.24 in the sale, what di

yawa3891 [41]

Answer:162.72

Step-by-step explanation:

Let's take the final price as x

What has been said is that before the sale, the price had been 3x, since the price would be three time higher. After the sale we divide the price by 3 and get price in the question. So, 53.24 multiplied by three is equal to the initial price before the sale. 1/3, just means that everything is cut off 3 times. You can prove this be adding 54.24, the price now, with two times itself or just multiplying 54.24 by three.

7 0

3 years ago

The table below shows how much Joe earns, y, after working x hours.

allochka39001 [22]

Both slopes have the same value which is 7.5.

To compute for the slope we will use this formula:
m = y^2 - y^1 over x^2 - x^1

For the first slope, the points are<span> (4, 30) and (12, 90). Substitute and simplify.
m = (90 - 30) over (12 - 4)
m = 7.5

For the second slope, the points are </span><span>(4, 30) and (10, 75). Same with the above computation.
m = (75 - 30) over (10 - 4)
m = 7.5

Since the relationship is linear, the slope does not change. </span>

6 0

3 years ago

Read 2 more answers