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3 years ago

Examine Tamika's work below. What is Tamika's error? 18a^-5b^-6/30a^3b^-5=3a^-2b^-11/5

Mathematics

2 answers:

Phantasy [73]3 years ago

4 0

Answer:

answer is a on econ

Step-by-step explanation:

vlada-n [284]3 years ago

3 0

Answer:

She added the exponents

Step-by-step explanation:

:) hope that helps

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.John has $40. He will use some of the money to buy books, and he wants to have at least $10 left to put in his savings account.

Crank

Answer:

The required inequality is $40-6b\leq 10$ .

Jhon can buy at most 5 books.

Step-by-step explanation:

Consider the provided information.

Part(A)

Let b represent the number of books John buys.

He wants at least $10 in his savings account which means if we subtract the cost of books from $40 we have at least $10 left.

$40-6b\leq 10$

The required inequality is $40-6b\leq 10$ .

Part(B)

$40-6b\leq 10$

$-6b\leq 10-40$

$b \leq\frac{30}{6}$

$b\leq 5$

Part(C): Jhon can buy at most 5 books.

3 0

3 years ago

100*.005<br><img src="https://tex.z-dn.net/?f=100%20%5Ctimes%20.005" id="TexFormula1" title="100 \times .005" alt="100 \times .0

andrew11 [14]

Answer is .5
move decimal place 2 places right

7 0

3 years ago

1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals n

Andrew [12]

Part 1
We are given $x^2-21x=-4x$ . This can be rewritten as $x^2-18x=0$ .
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}$
$x=\frac{18\pm 18}{2}$

The values of x are
$x_1=\frac{18-18}{2}=0$
$x_2=\frac{18+18}{2}=18$

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.

Part 3.
The first equation is $y=x^2+2$ .
The second equation is $y=3x+20$ .

We have
$x^2+2=3x+20$
$x^2-3x-18=0$
Factoring, we have
$\left(x-6\right)\left(x+3\right)=0$
Equating both factors to zero.
$x_1-6=0\rightarrow x_1=6$
$x_2+3=0\rightarrow x_2=-3$

When the value of x is 6, the value of y is
$y=3\left(6\right)+20=38$

When the value of x is -3, the value of y is
$y=3\left(-3\right)+20=11$

Therefore, the solutions are (6,38) or (-3,11)

7 0

2 years ago

On a coordinate plane, a line goes through (negative 4, 0) and (0, 2).

Kazeer [188]

Answer:

A C D

Step-by-step explanation:

3 0

2 years ago

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What is the range of this function?

pantera1 [17]