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777dan777 [17]
3 years ago
14

HELP!!!

Mathematics
1 answer:
lesya [120]3 years ago
6 0

Answer:

its D

Step-by-step explanation:

basically all you have to do if cut off the semicircles fins their areas then find the rectangles area and add them all up

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When x = 5, the value of the expression –2(x – 10) is
nikitadnepr [17]

Answer:

The value of the expression at x=5 is 10

Step-by-step explanation:

The given expression is -2(x-10)

Let us simplify this expression.

Distribute -2 over the parentheses, we get

-2x+20

Now, substitute x = 5, we get

-2(5)+20

On simplifying, we get

-10+20\\=10

The value of the expression at x=5 is 10

8 0
3 years ago
Read 2 more answers
What is the domain and range for:<br> y = arcsin(tanx)
Semmy [17]
The graph looks like this, on the enclosed pic:
One feature is that it's periodic and torn (has cut-off points), meaning the domain is the same as in case of tan(x): x€R and x =/= π/2+πn.
The range equals the range of arcsin(x): -π/2<=y<=π/2 OR y€[-π/2;π/2]
Hope could understand and if it helped! :)

7 0
3 years ago
B-4/6 how do I solve this? It's hard
serious [3.7K]
                                     
if you are looking for.... 2(B-4)= (6)b
                                     2b - 8 = 6b
                                     2b - 6b= 8
                                     -4b=8
                                      so b= -2

the question is b-4 divided by 6 equals b/2

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

6 0
2 years ago
I need help with this question
mariarad [96]

Answer:

133

Step-by-step explanation:

Answer is 133

6 0
3 years ago
Read 2 more answers
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