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3 years ago

Need help im struggling

Mathematics

1 answer:

alexira [117]3 years ago

3 0

I believe the answer will be 23in

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If f(3)=5 and f(4)=8 what does f(5)= and f(6)=

Novosadov [1.4K]

They will equal 11 and 14 it moves up by 3... Did you know that when you walk you move................................MIND-BLOWN

4 0

3 years ago

Pls help me in this i really need help :( :)

Andrews [41]

Hello and Good Morning/Afternoon:

Let's take this problem step-by-step:

First off, let's write the line in point-slope form:

$\rm \hookrightarrow (y-y_0)=m(x-x_0)$

(x₀, y₀) any random point on the line
'm' is the value of the slope

Let's calculate the slope:

$\rm \hookrightarrow Slope = \frac{y_2-y_1}{x_2-x_1}$

(x₁, y₁): any random point on the line ⇒ (-2, -6)
(x₂,y₂): any random point on the line that is not (x₁, y₁) ⇒ (2, -3)

$\rm \hookrightarrow slope = \frac{-3--6}{2--2} =\frac{3}{4}$

Now that we found the slope, let's put it into the point-slope form

⇒ we need (x₀, y₀) ⇒ let's use (2,-3)

$(y-(-3))=\frac{3}{4} (x-2)\\y+3=\frac{3}{4} (x-2)$

The equation, however, could also be put into 'slope-intercept form'

⇒ gotten by isolating the 'y' variable to the left

$y = \frac{3}{4}x-\frac{9}{2}$

Answer: $y = \frac{3}{4}x-\frac{9}{2}$ or $y+3=\frac{3}{4} (x-2)$

*Either equations work, put the one that you are the most familiar with

Hope that helps!

#LearnwithBrainly

5 0

2 years ago

Y log c - 8 log r Condense the logarithm.

HACTEHA [7]

Answer:

log[(c^y)/(r^8)] this is m your answer

6 0

3 years ago

A building company bids on two large projects. The CEO believes the chance of winning the 1st is 0.6, the chance of winning the

Ksju [112]

Answer:

0.8

Step-by-step explanation:

A: P(1) = .6, P(2) = .5, P(1 and 2) = .3

P(1 or 2) = P(1) + P(2) - P(1 and 2) .6 + .5 - .3 = 0.8

4 0

3 years ago

Write an equation for each line

Semmy [17]

There is no information here, could you add more. There are no lines for us to help you with so this makes it confusing for others, add more information for us to be able to help.

4 0

3 years ago

Need help im struggling​

Need help im struggling