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2 years ago

How do you write 4.5 × 102 in standard form?

Mathematics

2 answers:

Gnoma [55]2 years ago

7 0

Step-by-step explanation:

4,5 x 102?

$4.5 \times 102 \\ = \frac{45}{10} \times 102 \\ = \frac{4590}{10} \\ = 459$

<h2>The answer is 459 </h2>

Basile [38]2 years ago

3 0

$\blue{ \boxed{ \tt{} \bold{ \: answer \: by : \boxed{ \green \star{ \tt{} \bold{ \red{ \: febryantohutagalung \: }} \green \star}}}}} \\ \\$

Answer:

$\tt{}459 \\ \\$

Step-by-step explanation:

$\tt{}4.5 \times 102 \\ \\ \tt{} \frac{45}{10} \times \frac{102}{1} \\ \\ \tt{} \frac{45 \times 102}{10 \times 1} \\ \\ \tt{} \frac{4590}{10} \: \: \: \: \: \: \: \: \\ \\ \tt{}459 \: \: \: \: \: \: \: \: \: \\ \\ \\ \\$

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If m∠XWY = 20°, m∠XWZ = 40°, and XY = 16, what is the value of YZ?

Jobisdone [24]

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2 years ago

Match each value with its formula for ABC.

Ivahew [28]

For the triangle ABC use the sine theorem:

$\dfrac{a}{\sin A}= \dfrac{b}{\sin B}= \dfrac{c}{\sin C}$ .

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4. From $\dfrac{a}{\sin A}= \dfrac{c}{\sin C}$ you have $\dfrac{\sin A}{\sin C} \cdot c=a$ .

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3 years ago

Read 2 more answers

Given that y=x^(1/3) , use calculus to determine an approximate value for ∛8030

Alex777 [14]

F(x) = x^(1/3)
f(8000) = 8000^(1/3) = 20

f'(x) = 1/3 * x^(-2/3)
f'(x) = 1/(3x^(2/3))
f'(8000) = 1/(3*8000^(2/3)) = 1/(3*400) = 1/1200

Linear approximation for f(x) about x = 8000
L(x) = f(8000) + f'(8000) (x−8000)
L(x) = 20 + 1/1200 (x−8000)

8030^(1/3) ≈ L(8030)
= 20 + 1/1200 (8030−8000)
= 20 + 30/1200
= 20 + 1/40
= 20.025

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2 years ago

A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A

julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

$D^{2} = (90 - x)^{2} + 90^{2}$ (1)

$\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}$

$2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}$

$D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}$ (2)

Since:

$D = \sqrt{(90 -x)^{2} + 90^{2}}$

When x = 45 (the batter is halfway to first base), D is:

$D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62$

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

$100.62 \frac{d(D)}{dt} = -(90 - 45)*31$

$\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s$

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

$D^{2} = 90^{2} + x^{2}$