All categories

3 years ago

Please help me with this homework

Mathematics

1 answer:

yulyashka [42]3 years ago

5 0

B= -42 C= -38
So the difference going up to positive is 4 and the difference going down in negative is -4.

You might be interested in

A school assembly had 70 students in attendance, and 10% of them were first-graders. How many first-graders were at the assembly

laila [671]

7 of them were first graders.

Find how many first graders by using the percentage of total students and number of total students. Multiply them together, and 10% of 70 students is 7 students, which are the first graders.

3 0

3 years ago

Read 2 more answers

Please solve the question below 2.0

Alexus [3.1K]

Answer:

The answer is B

Step-by-step explanation:

Just look at the picture

3 0

3 years ago

Complete the solution of the equation. Find the

PolarNik [594]

Answer:

y = 7

Step-by-step explanation:

hope this helps

6 0

3 years ago

The Pear company sells pPhones. The cost to manufacture x pPhones is C ( x ) = - 21 x² + 67000 x + 20006 dollars (this includes

pishuonlain [190]

Answer:

Company should produce 8000 phones for the maximum profit.

Step-by-step explanation:

Function representing cost to the manufacturer is,

C(x) = -21x²+ 67000x + 20006

Function that represents the revenue generated is,

R(x) = -28x² + 179000x

Function representing profit to the company,

Profit = Revenue generated - Cost price

= -28x² + 179000x - (- 21x² + 67000x + 20006)

= -28x² + 21x² + 179000x - 67000x - 20006

P(x) = -7x² + 112000x - 20006

Since this function is a quadratic function therefore, the maximum price generated will be for

$x=-\frac{b}{2a}$

From the given quadratic function, a = (-7), b = 112000 and c = -20006

Therefore, for $x=-\frac{112000}{2\times (-7)}$

Or x = 8000 the profit to the company will be maximum.

That means pear company should produce 8000 phones for the maximum profit.

7 0

3 years ago

Solve this algebra...

Mumz [18]

Step-by-step explanation:

$\text{Use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ (a^n)^m=a^{nm}\\\\\left(\dfrac{a^x}{a^y}\right)^{x-y}=\left(a^{x-y}\right)^{x-y}=a^{(x-y)(x-y)}=a^{(x-y)^2}\\\\\left(\dfrac{a^y}{a^z}\right)^{y-z}=\left(a^{y-z}\right)^{y-z}=a^{(y-z)(y-z)}=a^{(y-z)^2}\\\\\left(\dfrac{a^z}{a^x}\right)^{z-x}=\left(a^{z-x}\right)^{z-x}=a^{(z-x)(z-x)}=a^{(z-x)^2}\\\\\text{Use}\ a^n\cdot a^m=a^{n+m}\\\\a^{(x-y)^2}\cdot a^{(y-z)^2}\cdot a^{(z-x)^2}=a^{(x-y)^2+(y-z)^2+(z-x)^2}$

$\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-y)^2=x^2-2xy+y^2\\(y-z)^2=y^2-2yz+z^2\\(z-x)^2=z^2-2zx+x^2\\\\=a^{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\\\\\text{combine like terms}\\\\=a^{(x^2+x^2)+(y^2+y^2)+(z^2+z^2)-2xy-2yz-2zx}\\\\=a^{2x^2+2y^2+2z^2-2xy-2yz-2zx}\qquad\text{distributive}\\\\=a^{2(x^2+y^2+z^2)-2(xy-yz-zx)}\\\\\text{From the equastion we know:}\ x^2+y^2+z^2=xy+yx+zx.\\\text{Therefore}\\\\=a^{2(x^2+y^2+z^2)-2(x^2+y^2+z^2)}=a^0=1$

6 0

4 years ago