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Nadya [2.5K]
3 years ago
12

Someone help me... please show work​

Mathematics
1 answer:
frutty [35]3 years ago
7 0

Answer:

x=5

Step-by-step explanation:

Set the 2 equations equal to each other and solve for x.

11x-28=7x-8

add 28 to both sides (-8+28=20)

11x=7x+20

subtract 7x from both sides (11x-7x=4x)

4x=20

divide both sides by 4 to isolate x (20/4=5)

x=5

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Which pair of angles is complementary?
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Answer:

2 angles are shown one angle is 56 degrees and the other is 34 degrees

Step-by-step exp

complementary angle=90 degree

56+34=90

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3 years ago
Of the 800 people at the concert, 42% of them are teenagers. How many teenagers are at the concert?
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Answer:

There are 336 teenagers in the concert. Please Mark me brainliest if I am correct thank you and have a nice day.

Step-by-step explanation:

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3 years ago
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Which of the following lists does not include all the factors of the number?
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The Answer is 21 because 7 and 3 can go in it as well
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3 years ago
In 24 hours one set of twins drank 16 bottles of formula. Each bottle holds 4 fl oz. How many cups of formula did the set of twi
aev [14]

Answer:

16 cups of formula

Step-by-step explanation:

you take 16 (<em>the amount of bottles)</em> and multiply it by 4 (<em>the amount one bottle holds</em>)

which gets you 64 then you multiply that by 2 (<em> the amount of children</em>)

which gets you 128

then you divide by 8, because there are <em>8 fl oz in a cup</em>

which gets you 16

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7 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
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Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
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