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2 years ago

Ashlee has finished 7/25 of her homework what percent of the homework does Ashlee still need to finish

Mathematics

2 answers:

kumpel [21]2 years ago

6 0

She has done 15% of her homework she needs 90 percent of her homework

zepelin [54]2 years ago

3 0

She has 18% of her homework to do.

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Over a four-week period, Gail earned the following commissions:

stellarik [79]

Answer:$257.5

Step-by-step explanation: add all the amounts then divide by 4

5 0

2 years ago

Read 2 more answers

Find the area bounded by the graphs of the indicated equations over the given interval.

NeX [460]

Answer:

Step-by-step explanation:

$y = 0$

is basically the horizontal axis.

First, find the integral of x^2-25.

Remember that

integral of a constant is that constant times x.

Also that

to take the integral of a power function, add 1 to the degree and divide by that same degree.

$y = {x}^{2} - 25$

We then get

$\frac{ {x}^{3} }{3} - 25x$

Evaluate at -3

$\frac{ - 3 {}^{3} }{3} - 25( - 3)$

$- 27 + 75 = 48$

Then we evaluate at 0

$\frac{0 {}^{3} }{3} - 25(0) = 0$

Next, we subtract the the answer then we get

$48 - 0 = 48$

8 0

2 years ago

Simplify 2(a-3)/(a-4)(a-5)+(a-1)/(3-a)(a-4)+(a-2)/(5-a)(a-3)

Umnica [9.8K]

Answer:

$\dfrac{5}{x^3-12x^2+47x-60}$

Step-by-step explanation:

Though it is not what you have written, we think you want to simplify ...

$\dfrac{2(a-3)}{(a-4)(a-5)}+\dfrac{(a-1)}{(3-a)(a-4)}+\dfrac{(a-2)}{(5-a)(a-3)}\\\\=\dfrac{2(a-3)^2}{(a-3)(a-4)(a-5)}+\dfrac{-(a-1)(a-5)}{(a-3)(a-4)(a-5)}+\dfrac{-(a-2)(a-4)}{(a-3)(a-4)(a-5)}\\\\=\dfrac{2(a^2-6a+9)-(a^2-6a+5)-(a^2-6a+8)}{x^3-12x^2+47x-60}\\\\=\boxed{\dfrac{5}{x^3-12x^2+47x-60}}$

_____

When writing a fraction in plain text, parentheses are needed around the entire denominator. The order of operations tells you that a/bc = (a/b)c, not a/(bc).

6 0

2 years ago

Write equations for the vertical and horizontal lines passing through the point (-4,-9).

Gnesinka [82]

Answer:

Below.

Step-by-step explanation:

These are perpendicular line which intersect at (-4, -9).

Vertical line is x = -4.

Horizontal lone is y = -9.

6 0

2 years ago

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago