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3 years ago

Given that Flx) = x2 + 2, evaluate F(2).

Mathematics

1 answer:

wariber [46]3 years ago

3 0

Step-by-step explanation:

f(2) = 2^2 +2

= 4+2

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The right triangle on the right is a scaled copy of the right triangle on the left. Identify

hodyreva [135]

Answer:

scale factor : 3.5

Step-by-step explanation:

56/16=3.5

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2 years ago

3.1 .31 1.3 .13 from least to greatest

katen-ka-za [31]

Hello there!

The answer is:
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8 0

3 years ago

3.8*10 to the 9th power divided by 4*10 to the 2nd power

user100 [1]

Answer:

$9.5\times 10^{6}$

Step-by-step explanation:

1. Divide the coefficients and the exponentials separately

$\dfrac{3.8 \times 10^{9}}{4 \times 10^{2}} = \dfrac{3.8}{4} \times \dfrac{10^{9}}{10^{2}}$

2. Divide the coefficients

$\dfrac{3.8}{4} = 0.95$

3. Divide the exponentials

Subtract the exponent in the denominator from the exponent in the numerator.

$\dfrac{10^{9}}{10^{2}} = 10^{(9 - 2)} = 10 ^{7}$

4. Re-join the new coefficient and the new exponential

$\dfrac{3.8 \times 10^{9}}{4 \times 10^{2}} = 0.95 \times 10^{7}$

5. Put the new number into standard form

The number before the power of 10 must be greater than or equal to one and less than 10.

Multiply the answer by 10/10.

$(0.95 \times 10^{7}) \times \dfrac{10}{10} = (0.95\times10) \times \dfrac{10^{7}}{10} = \mathbf{9.5\times 10^{6}}$

5 0

3 years ago

Write the quadratic equation the has the roots (-1-√2)/3 and (-1+√2)/3 if its coefficient with x^2 is equal to...1

DerKrebs [107]

Step-by-step explanation:

x1+x2 = (-1-√2)/3 + (-1+√2)/3

= -2/3

(x1) (x2) = (-1-√2)/3 × (-1+√2)/3

= -1/9

the equation :

x² -(-2/3)x + (-1/9) = 0

x² + ⅔ x - 1/9 = 0

5 0

3 years ago

From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$

Hence, the mat is 4.33 ft high off the ground.

5 0

3 years ago

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