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2 years ago

11 less than three times a number is the same as 75 less than the product of -1 and the number

Mathematics

2 answers:

Elina [12.6K]2 years ago

5 0

Inequality?

11 < 3x = 75 < -1x

bekas [8.4K]2 years ago

4 0

11 less than (-11) 3 times a number (3n) is the same as (=) 75 less than (-75) the product of -1 and the number (-1 times n or -n)

-11+3n=-75-n
solve
add n to both sides
-11+4n=-75
add 11 to both sides
4n=-64
divide both sides by 4
n=-16

the number is -16

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While walking in her neighborhood, Ashley found 3 dimes and 4 pennies on the ground. What percent of a dollar did Ashley find?

faltersainse [42]

Answer:

34%

Step-by-step explanation:

Dime = 10 cents, 10% of a dollar

Penny = 1 cent, 1% of a dollar

3 dimes = 30% of a dollar

4 pennies = 4% of a dollar

30% + 4% = 34%

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2 years ago

Please answer step by step: -4=2x+16 What does x equal? x=____?

Fed [463]

Answer:

Step-by-step explanation:

-4=2x+16

-16. -16

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20/2=2x/2

10=x

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2 years ago

Pls help me find perimeter and area, id really appreciate it. thank u!

Agata [3.3K]

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2 years ago

What is the slope of the line 7, 0 and -5, 0

lutik1710 [3]

Answer:

Slope = (Y2 -Y1) ÷ (X2 -X1)

Slope = (0 -0) / (-5, -7)

Slope = 0 / -12 = 0

The slope of the line is zero. It is perfectly horizontal.

Step-by-step explanation:

6 0

2 years ago

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

Answer:

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

Solution:

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

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3 years ago