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3 years ago

What is 5.4 x 2 3/4 show your work

Mathematics

2 answers:

VladimirAG [237]3 years ago

3 0

Answer:

14.85

Step-by-step explanation:

4.4 x 2.75 = 14.84

2.75

4.4

-------

(Multiply it out)

Ratling [72]3 years ago

3 0

Answer:

14.85

Step-by-step explanation

5 * 2 = 10

0.4 * 2 = 0.8

5 * 0.75 (0.75 = 3/4) = 3.75

0.4 * 0.75 = 0.3

10 + 0.8 = 10.8 + 3.75 = 14.55 + 0.3 = 14.85

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Please help solve this system of equations

stepan [7]

Make a substitution:

$\begin{cases}u=2x+y\\v=2x-y\end{cases}$

Then the system becomes

$\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}$

Simplifying the equations gives

$\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}$

which is to say,

$\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}$

$\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81$

$\implies\dfrac uv=\pm27$

$\implies u=\pm27v$

Substituting this into the new system gives

$\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1$

$\implies u=\pm27$

Then

$\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13$

(meaning two solutions are (7, 13) and (-7, -13))

8 0

3 years ago

The sum of the measures of two complementary angles is 90°. If one angle measures 24° more than twice the measure of the other,

geniusboy [140]

X + y = 90

Let's make 'x' the bigger angle and 'y' the smaller.
X is 24 more than twice y.
x = 24 + 2y

Now plug that into the x of the first equation.
(24 + 2y) + y = 90

Combine like terms.
24 + 3y = 90

Subtract 24 from both sides.
3y = 66

Divide both sides by 3.
y = 22

The measure of the smaller angle is 22°

3 0

3 years ago

A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate

Lyrx [107]

Answer:

Therefore the rate change of height is $\frac{1}{\pi}$ m/s.