All categories

2 years ago

You have 8 dimes. You trade each dime for 10 pennies. How many pennies will you have?

Mathematics

2 answers:

Salsk061 [2.6K]2 years ago

8 0

You will have 80 pennies
if I dime is 10 pennies then
8×10=80 pennies

Kamila [148]2 years ago

3 0

Answer:

you will have 80 pennies

Step-by-step explanation:

8*10=80

You might be interested in

Find all solutions to the equation in the interval [0, 2īt).<br> 4) sin 2x = -sin x

Sonbull [250]

Answer:

x={0, 2pi/3, pi, 4pi/3}

Step-by-step explanation:

First, move sin x by adding sin x to both sides. sin 2x+ sin x = 0. Next using the double angle identity, sin 2x=2(sin x)(cos x), so 2(sin x)(cos x)+sinx=0. Factoring, sin x(2cos x +1)=0. Solving, sin x=0 and 2cos x +1 = 0, or cos x = -1/2.

sin x = 0. Using the unit circle, sin x=0 when x=0 and x=pi.

cosx=-1/2. Using the unit circle, cos x= -1/2 when x=2pi/3 and x=4pi/3.

8 0

2 years ago

Help me with questions 8 and 9 for 20 POINTS!!!!!!

rewona [7]

Answer:

hehejejejjenene

Step-by-step explanation:

sjjshenebjsjsjwjjejenene .....

hope it helps

5 0

2 years ago

Simplify: 9(1 - r) + 3r

Butoxors [25]

Answer:

hdudhryhrgrhdidbf7hs6he8hs6jw9wg5gssi

4 0

2 years ago

Read 2 more answers

Select the correct location in the expression.

kogti [31]

Answer:

Correct term:60

Step-by-step explanation:

We are given that

$\frac{4x^4-6x^3+6x+3}{x-3}$

The quotient of this expression is given by

$4x^3+6x^2+18x+78+\frac{183}{x-3}$

We have to find the term in this expression which contains an error.

$\frac{4x^4-6x^3+6x+3}{x-3}$

The quotient of this expression is given by

$=4x^3+6x^2+18x+60$

$Dividend=Divisor\times quotient+ remainder$

Using the formula

$4x^4-6x^3+6x+3=(x-3)(4x^3+6x^2+18x+60)+\frac{183}{x-3}$

We can see that in the given expression there is 60 in place of 78.

The error in the expression of quotient is 78.

Correct term is 60.

4 0

2 years ago

Simplify the expression

kow [346]

Answer:

Step-by-step explanation:

$\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}\\\\6(x^2-3x+2)\ne0\ \iff\ x=\frac{3\pm\sqrt{9-8}}{2}\ne0\ \iff\ x\ne2\ \wedge\ x\ne1\\\\\\\dfrac{5x^3-8x^2-4x}{6x^2-18x+12}=\dfrac{x(5x^2-8x-4)}{6(x^2-3x+2)}=\dfrac{x(5x^2-10x+2x-4)}{6(x^2-2x-x+2)}=\\\\\\=\dfrac{x[5x(x-2)+2(x-2)]}{6[x(x-2)-(x-2)]} =\dfrac{x(x-2)(5x+2)}{6(x-2)(x-1)}=\dfrac{x(5x+2)}{6(x-1)}=\dfrac{5x^2+2x}{6x-6}\\\\\\f(x)=5x^2+2x\\\\g(x)=6x-6$

4 0

2 years ago