All categories

3 years ago

4/5 x 3 x ½ = I need this to pass, thanks.

Mathematics

2 answers:

dsp733 years ago

4 0

Answer:

4/5 x 3 x ½ = $\frac{6}{5}$ or $1\frac{1}{5}$

Step-by-step explanation:

You can set up your given question by transforming all of the numbers into fraction:

$\frac{4}{5} *\frac{3}{1} *\frac{1}{2}$

Then, you can multiply the numerators together, and the denominators:

$\frac{4*3*1}{5*1*2} = \frac{12}{10}$

Reducing $\frac{12}{10}$ down to lowest terms will give you: $\frac{12}{10}$ ÷ $\frac{2}{2}$ = $\frac{6}{5}$ or $1\frac{1}{5}$

Y_Kistochka [10]3 years ago

3 0

Answer:

i hope this helps , mark me brainlist

You might be interested in

How to write a fraction or mixed number as a decimal pre algebra?

Oksanka [162]

If you had the number 4 1/4 you would divide the numerator (1) by the denominator (4) and then take the decimal number (0.25) you get and add the decimal to the whole number on your mixed number (4+0.25) and that would give you the decimal.

4 0

3 years ago

1.) Write an equation of the line that passes through the point (3,-2) and is

GarryVolchara [31]

Answer:

y = 2x - 8

Step-by-step explanation:

8 0

3 years ago

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

8 0

3 years ago

What is the oblique asymptote of gx) = x2-3x-5?

oee [108]

Answer:

B) $y=x-5$

Step-by-step explanation:

Because $g(x)=\frac{x^2-3x-5}{x+2}$ simplifies to $x-5+\frac{5}{x+2}$ , we only care about the quotient, which will be our oblique asymptote equation. Therefore, the oblique asymptote for the function will be $y=x-5$ . See the attached graph for a visual.