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3 years ago

)A scanner scanned 63 photos in 9 minutes. If it scans photos at a constant rate, it can scan _____ photos in 21 minutes.

Mathematics

1 answer:

Cloud [144]3 years ago

3 0

Answer:

147 photos

Step-by-step explanation:

63 photos -> 9 minutes

To find the number of photos scanned per minute, divide both sides by 9.

$63\div 9=7$ photos -> 1 minute

To find 21 minutes, we can do $21\times 7=147$

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Lean wants to giftwrap a present she got for her little brother. how many square inches of giftwrap will be needed to cover a bo

Natalka [10]

Check the picture below.

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the front and back are two 5x3 rectangles

the left and right are just two 3x7 rectangles

and the top and bottom are just two 5x7 rectangles

so, just get the area of each, sum them up, and that's the area of the rectangular prism and thus of the box, and that's how many in² she will need.

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3 years ago

HELP ME!! I WILL GIVE BRAINLIEST!!! Express your answer in SCIENTIFIC NOTATION. 9.3 × 10⁷ - 3.4 × 10⁶ = __________ plz help!!

DanielleElmas [232]

Answer:

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3 years ago

F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte

Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

2x² - x -1 = 0

( x - 1) (2x + 1) = 0

x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

x² + 2 = 0

x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

$\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}$

The function is concave up on (0, ∞).

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4 years ago

If a triangle is an equilateral triangle, then the triangle has exactly three 60°

ahrayia [7]

Step-by-step explanation:

In an equilateral triangle, all the three sides are equal as well as all the angles are equal. Let the angles be x.

We know that the sum of angles of a triangle is equal to 180 degrees. It means that,

x+x+x=180

3x=180

x=60°

Hence, if a triangle is an equilateral triangle, then the triangle has exactly three 60° angles. Hence, △ABC is an equilateral triangle

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4 years ago

Which logarithmic equation is equivalent to the exponential equation below?

aleksandr82 [10.1K]

Option B: $\ln 6=5 x$ is the correct answer.

Explanation:

The exponential equation is $e^{5 x}=6$

If $f(x)=g(x)$ , then $\ln (f(x))=\ln (g(x))$

Thus, the equation becomes

$\ln \left(e^{5 x}\right)=\ln (6)$

Applying log rule, $\log _{a}\left(x^{b}\right)=b \cdot \log _{a}(x)$ and thus the equation becomes

$5 x \ln (e)=\ln (6)$

Since, we know that, $\ln (e)=1$ , using this we get,

$5 x=\ln (6)$

Hence, the logarithmic equation which is equivalent to the exponential equation $e^{5 x}=6$ is $\ln 6=5 x$

Thus, Option B is the correct answer.

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3 years ago