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2 years ago

If ABCD is a square and AB = 10, what is the measure of CE to the nearest hundredth?

Mathematics

1 answer:

andrey2020 [161]2 years ago

6 0

Answer:

$AC=14.14\ units$

Step-by-step explanation:

The correct question is

If ABCD is a square and AB = 10, what is the measure of AC? (rounded to the nearest hundredth)

see the attached figure to better understand the problem

we know that

All four sides of the square are congruent and all four interior angles are equal to 90°.

AB=BC=CD=AD

In the right triangle ACD

Applying the Pythagorean Theorem

$AC^2=CD^2+AD^2$

we have that

$CD=AD=AB=10\ units$

substitute the given value

$AC^2=10^2+10^2$

$AC^2=200$

$AC=\sqrt{200}\ units$

$AC=14.14\ units$

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Option c (Upper tailed) is the correct choice.

Step-by-step explanation:

Given that:

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2 years ago

Enrique is making a party mix that contains raisins and nuts. For each ounce of nuts, he uses twice the amount of raisins. How m

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2 years ago

Solve the system of equations – 5x + y = 56 and x + y = -4 by combining the equations.

anygoal [31]

Answer:

Your solution is (-10, 6).

Step-by-step explanation:

Combining the equations is also known as substitution. This is done when you substitute one variable into another equation.

-5x + y = 56

x + y = -4

Let's change the second equation into one with one variable on each side.

y = -x - 4

Now, plug this into your first equation.

-5x + (-x - 4) = 56

Distribute the + sign.

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Combine the like terms.

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Isolate x by dividing both sides by -6.

x = -10

Now plug this back into either equation.

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Add 10 to both sides to find y.

y = 6

Your solution is (-10, 6).

Check this by plugging in these values into the equation you have not checked yet.

-5(-10) + (6) = 56

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Your solution is correct.

Hope this helps!

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1 year ago

Read 2 more answers

R+5/mn=p solve for m

AnnZ [28]

Answer:

$\bold{m\ =\ \dfrac5{(p-r)n}}$

Step-by-step explanation:

$\bold{r+\dfrac5{mn}\ =\ p}\\\\ {}\quad-r\qquad-r\\\\{}\ \ \bold{\dfrac5{mn}\ =\ p-r}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\quad\bold{5\ =\ (p-r)^{_\times}(mn)}\\\\\div(p-r)\quad\div(p-r)\\\\{}\ \ \bold{\dfrac5{p-r}\ =\ mn}\\\\{}\quad \ \div n\quad\ \ \div n\\\\\bold{\dfrac5{(p-r)n}\ =\ m}$

If you mean (r+5)/mn then:

$\bold{\dfrac{r+5}{mn}\ =\ p}\\\\{}\ ^{_\times}(mn)\quad ^{_\times}(mn)\\\\{}\ \bold{r+5\ =\ pmn}\\\\\div(pn)\quad\div(pn)\\\\{}\ \ \bold{\dfrac{r+5}{pn}\ =\ m}$

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2 years ago

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