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2 years ago

PLS HELP marking brainliest 20-27 show ur work

Mathematics

1 answer:

Ivanshal [37]2 years ago

8 0

Answer:

-7?

Step-by-step explanation:

I mean, is the question 20 - 27 =? Or it's something else? Lol.

20 - 27 = -7, it's very straightforward cannot write it in more detailed way. When we substract a number from another and the first one is smaller the result is a negative number. Imagine someone sells up to 3 apples, you ask him to give you 5. He gives you all 3 of his so he has 0 but you force him to go and get another 2 for you. He lost another 2, so he has less than 0 now. He has -2 apples: 3 - 5 = -2.

Now, I don't know if this is the question or these are set of exercises you must solve and your picture just didn't upload. Let me know if you need help with anything else.

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Can somebody help me put the numbers in the 2 empty boxes.

Alja [10]

Answer:

I think its 4 and 5

Step-by-step explanation:

8 0

3 years ago

Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.

solong [7]

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is: $f(x)= \frac{9-x^2}{x^2-4}$

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

<u>a. (-infinity, -3) </u>

put k = -4 from the given interval

We get $f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get $f(-2.5)= \frac{9-(-2.5)^2}{(-2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get $f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0$

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get $f(2.5)= \frac{9-(2.5)^2}{(2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get $f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (3,∞)

7 0

3 years ago

PLEASE HELP

muminat

o.83

Step-by-step explanation:

Conversion a mixed number 3 1/

to a improper fraction: 3 1/4 = 3 1/

= 3 · 4 + 1/

= 12 + 1/

= 13/

To find new numerator:

a) Multiply the whole number 3 by the denominator 4. Whole number 3 equally 3 * 4/

= 12/

b) Add the answer from previous step 12 to the numerator 1. New numerator is 12 + 1 = 13

c) Write a previous answer (new numerator 13) over the denominator 4.

Three and one quarter is thirteen quarters

Conversion a mixed number 2 1/

to a improper fraction: 2 1/3 = 2 1/

= 2 · 3 + 1/

= 6 + 1/

= 7/

To find new numerator:

a) Multiply the whole number 2 by the denominator 3. Whole number 2 equally 2 * 3/

= 6/

b) Add the answer from previous step 6 to the numerator 1. New numerator is 6 + 1 = 7

c) Write a previous answer (new numerator 7) over the denominator 3.

Two and one third is seven thirds

Add: 13/

+ 7/

= 13 · 3/

4 · 3

+ 7 · 4/

3 · 4

= 39/

+ 28/

= 39 + 28/

= 67/

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