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3 years ago

PLEASE HELP I KNOW ITS EASY BUT I FORGOT!!!! PLEASE HELP!! Write -3/8 as a decimal.

Mathematics

2 answers:

raketka [301]3 years ago

7 0

Answer:

I got you bro

Step-by-step explanation:

-0.38

Ivan3 years ago

5 0

Answer:

-0.375

Step-by-step explanation:

You can write fractions as equations and solve for a decimal.

-3/8 = -3 ÷ 8

-3 ÷ 8 = -0.375

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a flower bed is in the shape of a triangle with one side twice the length of the shortest side and the third side is 19 feet mor

Neporo4naja [7]

Let's assume

shorter side as x

so, first side =2x

third side =x+19

now, we can find perimeter of triangle

perimeter of any triangle is

first side + shortest side + third side

so, we get

$P=2x+x+x+19$

we are given perimeter=187

so, we can set it to 187

and then solve for x

$187=2x+x+x+19$

$187=4x+19$

$4x+19-19=187-19$

$4x=168$

$x=42$

so,

shortest length is x=42 feet

First side =2x=2*42=84feet

third side =x+19=42+19=61 feet

so, dimensions are

first side is 42 feet

second side is 84 feet

third side is 61 feet................Answer

3 0

3 years ago

This is the last time its happened twice please i need the answer

slamgirl [31]

Answer:

I think the answer is F hope its right fingers crossed!

Step-by-step explanation:

3 0

3 years ago

What’s 4/5 of 3 And what’s 2/9 of 20

Karo-lina-s [1.5K]

Answer:

4/5 of 3 = 12/5

2/9 of 20 = 40/9

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A=pm+z solve for m<br> m=?

Leno4ka [110]

$A=pm+z\\\\pm+z=A\ \ \ \ |subtract\ "z"\ from\ both\ sides\\\\pm=A-z\ \ \ \ |divide\ both\ sides\ by\ "p"\\\\\boxed{m=\frac{A-z}{p}}$

4 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago