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3 years ago

Only one of the comparisons below is correct. Which is correct? What benchmark was used in your answer?

Mathematics

1 answer:

tiny-mole [99]3 years ago

5 0

Answer:

2/3<9/10; I used 3/4 as a benchmark.

Step-by-step explanation:

2/3<1/2; I used 1/2 as a benchmark.

2/3 = 0.(20/3) = 0.667

1/2 = 0.(10/2) = 0.5

So this is wrong, as 0.667 > 0.5.

1/2=3/5; I used 1/4 as a benchmark.

1/2 = 0.(10/2) = 0.5

3/5 = 0.(30/5) = 0.6

0.5 != 0.6, so this is wrong.

2/3<9/10; I used 3/4 as a benchmark.

2/3 = 0.(20/3) = 0.667

9/10 = 0.(90/10) = 0.9

So this is correct, as 0.667 < 0.9

3/4<2/3; I used 1/2 as a benchmark.

3/4 = 0.(30/4) = 0.75

2/3 = 0.(20/3) = 0.667

0.75 > 0.667, so this is wrong.

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HELP DUE LIKE RIGHT NOW

Nimfa-mama [501]

Answer:

The correct answer is x = 17.

Step-by-step explanation:

If EF bisects DEG, this means that angles DEF and angles FEG are congruent, and they each make up half of angle DEG.

Therefore, we can set up the equation:

DEF + FEG = DEG

However, since we know that DEF and FEG represent the same value, we can change this equation into the following:

2(DEF) = DEG

Now, we can substitute in the expressions that we are given:

2(3x+1) = 5x + 19

To simplify, we should first use the distributive property on the left side of the equation.

6x + 2 = 5x + 19

Our next step is to subtract 5x from both sides of the equation.

x + 2 = 19

Finally, we can subtract 2 from both sides of the equation to get x by itself on the left side.

x = 17

Therefore, the value of x is 17.

Hope this helps!

7 0

4 years ago

Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0

3 years ago

Find the measure of ∠C to the nearest degree.

Alisiya [41]

Well, sine is $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}$ ,

but from the angle C, the opposite side, the one facing it off, is the side
AB, and we dunno what the side AB is

however, we know the hypotenuse is 8, and the adjacent side is 7,

and is right-triangle, thus, let us use the pythagorean theorem,

$\bf c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite
\end{cases}
\\\\
thus
\\\\
\sqrt{8^2-7^2}=b=AB$

now, that we know what the opposite side is, that is AB, then
we can find the sine of angle C $\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin(C)=\cfrac{\overline{AB}}{\overline{BC}}$