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3 years ago

PLZ HELP ASAP WILL MARK BRAINIEST (x−5)/2 + (x−1)/8 = (x−3)/4 + (x−4)/3 solve for x!!

Mathematics

1 answer:

Aliun [14]3 years ago

8 0

Answer:

x= 13

Step-by-step explanation:

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Given AB is a diameter, find the measure of

Sonbull [250]

Answer:

Angle BCA = 90

Step-by-step explanation:

A triangle inscribed inside of a circle using the diameter as it's hypotenuse is a right triangle

Where the angle opposite of the hypotenuse ( or diameter ) is the right angle

Angle BCA is the angle opposite of the diameter

Hence BCA = 90

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3 years ago

There are 10 girls and 8 boys in Mr Augellis class. If 5 more girls and 5 more boys join the class, will the ratio of girls to b

ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

I'm not quite sure what a ratio table is, but I can explain it the way I understand it.

The initial ratio is 10:8 which is 1.25 as a decimal.

Now let's add five to each number to represent the kids joining the class, this will result in the following:

15:13 which is 1.153846154 as a decimal.

Therefore, the ratio doesn't stay the same just because you added the same number of people to each group.

Hope this helps :)

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4 years ago

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rewona [7]

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5 0

3 years ago

Read 2 more answers

Pls help me on this

Nataly [62]

Answer:c=20r(2h)

Step-by-step explanation:Simple rate of change

8 0

3 years ago

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

IRISSAK [1]

Answer:

There is one extraneous solution at $x = 4$

The only solution is $x=7$

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sqrt{x-3}+5=x$

In order to solve, we will first isolate the radical part.

Adding -5 on both the sides, we get:

$\sqrt{x-3}+5-5=x-5\\\sqrt{x-3}=x-5$

Now, squaring both the sides, we get:

$(\sqrt{x-3})^2=(x-5)^2\\\\x-3=x^2+25-10x\\\\x^2-10x-x+25+3=0\\\\x^2-11x+28=0\\\\(x-4)(x-7)=0\\\\x=4\ or\ x=7$

Now, let us check whether the calculated values satisfy the equations or not.

Plug in 4 and 7 for 'x' and check the value of the equation.

At $x=4$ ,

$\sqrt{4-3}+5=4\\\sqrt1+5=4\\1+5=4\\6=4$

But, 6 is not equal to 4 and hence 4 is not a solution to the given equation. It is an extraneous solution for the given equation.

Now, at $x=7$

$\sqrt{7-3}+5=7\\\sqrt4+5=7\\2+5=7\\7=7(True)$

Therefore, $x=7$ is a solution to the equation.

Hence, there is one extraneous solution at $x = 4$

The only solution is $x=7$

6 0

4 years ago