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Slav-nsk [51]
3 years ago
15

A function is defined by f(x)=5(2-x). What is f(–1)?

Mathematics
1 answer:
liq [111]3 years ago
5 0
In the given function f(x) = 5(2-x), subst. -1 for x, obtaining f(-1) = 5(2-[-1])= 10.
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Using numbers 3 plus 4 times 5 minus 6 divied by 2 equal 32<br><img src="https://tex.z-dn.net/?f=3%20%2B%204%20%5Ctimes%205%20-%
stellarik [79]
31.

3+4 is 7

7 × 5 is 35

35-6 is 29

29 + 2 is 31
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4 years ago
150% of 60 and small explanation
gogolik [260]
We can turn 150% into a decimal by moving our decimal point two places to the left, or 1.5. We then multiply 1.5 with 60 as the word "of" in math automatically tells us to do that.

We have our answer of 90.
7 0
4 years ago
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A plane travels 2,000 kilometers at a speed of 900 kilometers per hour (kph) with no wind. when a tailwind is present, the plane
Nina [5.8K]

The meaning of the y-intercept exists in the time required for the plane to cover 2,000 kilometers at a speed of 900 kilometers per hour (kph) with no wind.  $2 \frac{2}{9}$ hours required for the plane to cover 2,000 kilometers at a speed of 900 kilometers per hour with no wind.

<h3>What is the meaning of the y-intercept for the function$t(x)=\frac{2,000}{900}+x $$?</h3>

The time it takes the plane to travel 2,000 kilometers with the tailwind, t(x), exists determined by the function

$t(x)=\frac{2,000}{900}+x $$

where x exists the increase in speed when the plane travels with the wind (tailwind).

The y-intercept exists at x=0, then

$&t(0)=\frac{2,000}{900}+x \\

$&t(0)=\frac{20}{9}=2 \frac{2}{9} hours

$2 \frac{2}{9}$ hours required for the plane to cover 2,000 kilometers at a speed of 900 kilometers per hour with no wind.

To learn more about y-intercept for the function refer to:

brainly.com/question/19552756

#SPJ4

3 0
2 years ago
Please Answer!! Will give brainly!!
shtirl [24]

Answer:

480 - 12.25d = 50

7 0
3 years ago
Evaluate the following integrals <br><br>​
marusya05 [52]

Answer:

a. (24 ln 2 − 7) / 9

b. x tan x + ln|cos x| + C

Step-by-step explanation:

a. ∫₁² x² ln x dx

Integrate by parts.

If u = ln x, then du = 1/x dx.

If dv = x² dx, then v = ⅓ x³.

∫ u dv = uv − ∫ v du

= (ln x) (⅓ x³) − ∫ (⅓ x³) (1/x dx)

= ⅓ x³ ln x − ∫ ⅓ x² dx

= ⅓ x³ ln x − ¹/₉ x³ + C

= ¹/₉ x³ (3 ln x − 1) + C

Evaluate between x=1 and x=2.

[¹/₉ 2³ (3 ln 2 − 1) + C] − [¹/₉ 1³ (3 ln 1 − 1) + C]

⁸/₉ (3 ln 2 − 1) + C + ¹/₉ − C

⁸/₉ (3 ln 2 − 1) + ¹/₉

⁸/₃ ln 2 − ⁸/₉ + ¹/₉

⁸/₃ ln 2 − ⁷/₉

(24 ln 2 − 7) / 9

b. ∫ x sec² x dx

Integrate by parts.

If u = x, then du = dx.

If dv = sec² x dx, then v = tan x.

∫ u dv = uv − ∫ v du

= x tan x − ∫ tan x dx

= x tan x + ∫ -sin x / cos x dx

= x tan x + ln|cos x| + C

8 0
4 years ago
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