All categories

3 years ago

May someone please help me with this ill give brainlyest if i figure out how to

Mathematics

2 answers:

Otrada [13]3 years ago

8 0

Answer: 5

Step-by-step explanation:

5x3+6=21

5x5=25

25-21=4

tresset_1 [31]3 years ago

7 0

Let "the number" = $n$

$6 + 3n = 5n - 4$

$6 + 3n - 5n = -4$

$6 - 2n = -4$

$-2n = -10$

$n = 5$

Check my work:

Six added to 3 times (5) is equal to 4 less than 5 times (5)

Six added to 15 is equal to 4 less than 25

21 is equal to 21

You might be interested in

Help me on this question 20÷10×10

puteri [66]

the answer is 20 because 20/10=2 and 2*10 is 20 so that is why 20 is your answer

7 0

3 years ago

Read 2 more answers

Which of the following graphs would represent the solution set for y > 4/5 x – 1?

vlada-n [284]

A>0 so function rises

b=-1 so for x=0 ⇒ y<-1

we look for y that is above the line

so answer is the last <span>graph.</span>

"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."

<span>Regards M.Y.</span>

7 0

4 years ago

Read 2 more answers

Cecilia packed a fruit basket for her new neighbors. If 20% of 30 pieces of fruit were bananas, what fraction of the fruit were

Anettt [7]

Answer:

6/30

Step-by-step explanation:

20 percent of 30 is 6 ^_^

7 0

3 years ago

Read 2 more answers

A new sidewalk will be 6 feet wide, 240 feet long, and filled to a depth of 6 inches (0.5 foot) with concrete. How many cubi

Andreyy89

6 * 240 * 0.5 = 720 feet³

720 feet³ =

26,6667 yards³ (your answer)

6 0

3 years ago

WILL MAKE BRAINLIEST!!!!Complete the system of equations with a linear equation so that the graphs of the equations intersect at

Nina [5.8K]

First, we have to find x- and y-intercepts of given equation.

1. y-intercept is a point (0, y) so substitute x = 0 and calculate:

$y=x^3-3x^2+x-3\\\\y= 0^3-3\cdot0^2+0-3\\\\y=0-0+0-3\\\\\boxed{y=-3}$

y-intercept = (0, -3)

2. x-intercept is a point (or points) (x, 0), so when we substitute y = 0 there will be:

$0= x^3-3x^2+x-3\\\\0=x^2(x-3)+(x-3)\\\\0=(x^2+1)(x-3)\\\\\\x^2+1=0\qquad\vee\qquad x-3=0\\\\x^2+1\text{ always}\ \textgreater \ 0\qquad\vee\qquad x=3\\\\\boxed{x=3}$

x-intercept = (3, 0)

Now, when we have x- and y-intercepts, we can write an equation of a line (graph of the linear function) that passes through this two points:

$A=(3,0)\qquad\qquad B=(0,-3)\\\\\\
y-y_A=\dfrac{y_B-y_A}{x_B-x_A}(x-x_A)\\\\\\y-0=\dfrac{-3-0}{0-3}(x-3)\\\\\\y=\dfrac{-3}{-3}(x-3)\\\\\\y=1(x-3)\\\\\\\boxed{y=x-3}$

Our linear equation is: y = x - 3

5 0

4 years ago