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3 years ago

5

WILL GIVE BRAINLIEST

1 answer:

Gre4nikov [31]3 years ago

3 0

Answer: slope: 1/4

y-intercept: 0,3

Step-by-step explanation:

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Solve the following algebraic equation using appropriate

Neko [114]

( x + 1 )/ 3 = x + 2
Multiply 3 on both sides
x + 1 = 3x + 6
Subtract 6 on both sides
x - 5 = 3x
Subtract x from both sides
-5 = 2x
Divide 2 from both sides
x = -2.5

6 0

3 years ago

A rectangular box has length 20cm, width 6cm and height 4cm. find how many cubes of size 2cm that will fit into the box.

Papessa [141]

<u>Answer:</u>

$\boxed{\pink{\sf The \ number \ of \ cubes \ that \ can \ be \ fitted \ is 60 .}}$

<u>Step-by-step explanation:</u>

Given dimensions of the box = 20cm × 6cm × 4cm .

Dimension of the cube = 2cm × 2cm × 2cm .

Therefore the number of cubes that can be fitted into the box will be equal to the Volume of box divided by the Volume of the cube. So ,

$\boxed{\red{\bf \implies No. \ of \ cubes \ = \dfrac{Volume \ of \ box}{Volume \ of \ cube }}}$

$\bf \implies n_{cubes} = \dfrac{20cm \times 6cm \times 4cm .}{2cm \times 2cm \times 2cm } \\\\\bf\implies n_{cubes} = 10 \ times 3cm \times 2cm \\\\\implies \boxed{\bf n_{cubes}= 60 }$

<h3><u>Hence</u><u> the</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>cubes</u><u> </u><u>that</u><u> </u><u>can</u><u> </u><u>be</u><u> </u><u>fitted</u><u> </u><u>in</u><u> the</u><u> </u><u>box </u><u>is</u><u> </u><u>6</u><u>0</u><u> </u><u>.</u></h3>

6 0

3 years ago

What is the equation of the line through (7, -1) with slope -3?

Slav-nsk [51]

Point slope is y-y1=m(x-x1) where m is slope and x1 and y1 are the given x and y values.

y-(-1)=-3(x-7)
y+1=-3(x-7)

Final answer: B

3 0

3 years ago

Read 2 more answers

Which of the following equations shows how substitution can be used to solve the following system of equations?

klasskru [66]

Answer:

i have no idea does anyone else know

Step-by-step explanation:

5 0

2 years ago

Give the properties for the equation -2x 2 - y + 10x - 7 = 0.

Sladkaya [172]

Given:

The quadratic equation is

$-2x^2-y+10x-7=0$

To find:

The vertex of the given quadratic equation.

Solution:

If a quadratic function is $f(x)=ax^2+bx+c$ , then

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

We have,

$-2x^2-y+10x-7=0$

It can be written as

$-2x^2+10x-7=y$

$y=-2x^2+10x-7$ ...(i)

Here, $a=-2,b=10,c=-7$ .

$\dfrac{-b}{2a}=\dfrac{-10}{2(-2)}$

$\dfrac{-b}{2a}=\dfrac{-10}{-4}$

$\dfrac{-b}{2a}=\dfrac{5}{2}$

Putting $x=\dfrac{5}{2}$ in (i), we get

$y=-2(\dfrac{5}{2})^2+10(\dfrac{5}{2})-7$

$y=-2(\dfrac{25}{4})+\dfrac{50}{2}-7$

$y=\dfrac{-50}{4}+25-7$

$y=\dfrac{-25}{2}+18$

On further simplification, we get

$y=\dfrac{-25+36}{2}$

$y=\dfrac{11}{2}$

So, the vertex of the given quadratic equation is $\left(\dfrac{5}{2},\dfrac{11}{2}\right)$ .

Therefore, the correct option is A.

3 0

3 years ago