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3 years ago

6

Sum of all digits for 754.75

1 answer:

andreyandreev [35.5K]3 years ago

3 0

700.00 + 50.00 + 4.00 + 0.7 + 0.05 = 754.75

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Can someone help me with these 2 questions?

Murljashka [212]

Which 2 questions theres no documents? if you tell me the 2 questions i can help:)

7 0

3 years ago

Shelly is going shopping at the mall to buy 3 pairs of shoes. She has a coupon for $2 off perpair of shoe after buying the first

aleksklad [387]

Answer:

$42

Step-by-step explanation:

If you want to check, simply take the values I calculated, multiply them by 3 and subtract them by 2 for the second and third pair, they will add up to the values that Shelly is willing to purchase, excluding tax.

8 0

3 years ago

Find an equation of the slant asymptote. Do not sketch the curve. y = 5x4 + x2 + x x3 − x2 + 5

murzikaleks [220]

Answer:

The slant asymptote is $y=5 x + 5$ .

Step-by-step explanation:

Line $y=mx+b$ is a slant asymptote of the function $y=f{\left(x \right)}$ , if either $m=\lim_{x \to \infty}\left(\frac{f{\left(x \right)}}{x}\right)=L$ or $m=\lim_{x \to -\infty}\left(\frac{f{\left(x \right)}}{x}\right)=L$ , and L is finite.

We want to find the slant asymptotes of the function

$f(x)=\frac{5 x^{4} + x^{2} + x}{x^{3} - x^{2} + 5}$

First, do polynomial long division

$\frac{5 x^{4} + x^{2} + x}{x^{3} - x^{2} + 5}=5 x + 5+\frac{6 x^{2} - 24 x - 25}{x^{3} - x^{2} + 5}$

Next, we use the above definition,

The first limit is

$\lim_{x \to \infty}\left(\frac{6x^2-24x-25}{x^3-x^2+5}\right)=0$

The second limit is

$\lim_{x \to -\infty}\left(\frac{6x^2-24x-25}{x^3-x^2+5}\right)=0$

The rational term approaches 0 as the variable approaches infinity.

Thus, the slant asymptote is $y=5 x + 5$ .

8 0

3 years ago

The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?

mars1129 [50]

Answer:

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

The sum of the squares of these roots is $\frac{-3}{2}$

Step-by-step explanation:

The given quadratic equation is $8x^2+12x-14$ has two real roots.

To find the roots .

$8x^2+12x-14=0$

Dividing the above equation by 2

$\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}$

$4x^2+6x-7=0$

For quadratic equation $ax^2+bx+c=0$ the solution is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a and b are coefficents of $x^2$ and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

$x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}$

$=\frac{-6\pm\sqrt{36+112}}{8}$

$=\frac{-6\pm\sqrt{148}}{8}$

$=\frac{-6\pm\sqrt{37\times 4}}{8}$

$=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}$

$=\frac{-6\pm\sqrt{37}\times 2}{8}$

$=2\frac{(-3\pm\sqrt{37})}{8}$

$=\frac{-3\pm\sqrt{37}}{4}$

$x=\frac{(-3\pm\sqrt{37})}{4}$

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

Now to find the sum of the squares of these roots

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}$

$=\frac{-6}{4}$

$=\frac{-3}{2}$

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}$

Therefore the sum of the squares of these roots is $\frac{-3}{2}$

3 0

3 years ago

If BPD and dpc have a sum<br>of 72 degrees What is APC?<br><br><br><br>A. 18°<br>B 90°<br>D.180°

quester [9]

The correct answer is : 108

7 0

3 years ago