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2 years ago

Explain how to fin 4 x 80. show your work

Mathematics

1 answer:

algol [13]2 years ago

5 0

4x80=320

Work:

4x8 = 32
(4+4+4+4+4+4+4+4)
Now carry the 0 and you get 320

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Which two values of x are roots of the polynomial below?

Julli [10]

Answer:

B. and C.

Step-by-step explanation:

1. Set the polynomial equal to zero:

x² + 5x + 7 = 0

2. Plug the given values of a, b, and c into the quadratic formula:

$x=\frac{-5+-\sqrt{5^{2}-4(1)(7) } }{2(1)}$

3. Solve the square root:

$x=\frac{-5+-\sqrt{25-4(1)(7)} }{2(1)}$

$x=\frac{-5+-\sqrt{25-28} }{2(1)}$

$x=\frac{-5+-\sqrt{-3} }{2(1)}$

4. Simplify the rest:

$x=\frac{-5+-\sqrt{-3} }{2}$

5. Separate the solutions:

$x=\frac{-5+\sqrt{-3} }{2} , x=\frac{-5-\sqrt{-3} }{2}$

Thus, options B and C are the answers.

hope this helps!

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2 years ago

How to multiply mixed fractions

morpeh [17]

Answer:

Step-by-step explanation:

Change each number to an improper fraction.

Simplify if possible.

Multiply the numerators and then the denominators.

Put answer in lowest terms.

Check to be sure the answer makes sense.

8 0

3 years ago

An engineer scale model shows a building that is 3 inches tall. If the scale is 1 inch = 600 feet, how tall is the actual buildi

Sever21 [200]

The actual height of building is 1800 feet.

Step-by-step explanation:

Given,

Height of building in model = 3 inches

Scale used by engineer;

1 inch = 600 feet

Therefore;

Actual height of building = Height in model * Scale used

Actual height of building = 3 * 600

Actual height of building = 1800 feet

The actual height of building is 1800 feet.

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2 years ago

Please help me <br><br> Question: What are the solutions to this graph?

Salsk061 [2.6K]

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2 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

2 years ago