All categories

3 years ago

Choose two number from 5 to 9. Use your numbers to write an addition sentence. Draw a picture to show your work

1 answer:

7 0

Bob has 8 apples and Sam gives him 6 more. Now he has 14 apples.
Now you just need to draw a picture of a guy with 8 apples and another guy giving him 6 more.

You might be interested in

What are the solutions of x^2+10x+21=0?

Luden [163]

(x+7)(x+3) so therefore you set each equation = 0...
x + 7 = 0
x + 3= 0
and solve
x = -3
x = -7

6 0

3 years ago

Read 2 more answers

Alex earns 12% commission on any of her monthly sales over $5,000. Her sales for May were $6,750. How much commission did Alex e

valentina_108 [34]

6750/100*12 = 810 is the answer

4 0

3 years ago

Which could be the first step in simplifying this expression? Check all that apply. (x cubed x Superscript negative 6 Baseline)

PtichkaEL [24]

Answer:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

Step-by-step explanation:

$(x^{3} x^{-6} )^{2}$ is the expression given to be solved.

First of all let us have a look at 3 formulas:

$1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}$

Both the formula can be applied to the expression( $(x^{3} x^{-6} )^{2}$ ) during the first step while solving it.

Applying formula (1):

$(x^{3} x^{-6} )^{2}$

Comparing the terms of $(x^{3} x^{-6} )$ with $p^a \times p^b$

$p=x, a =3, b=-6$

$\Rightarrow x^{3+(-6)}\\\Rightarrow x^{3-6}\\\Rightarrow x^{-3}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $(x^{-3} )^{2}$

Applying formula (2):

Comparing the terms of $(x^{3} x^{-6} )^{2}$ with $(p^a \times q^b)^c$

$p=q=x, a =3, b=-6, c=2$

$\Rightarrow (x^{3})^2\times (x^{-6})^2\\\text{Applying Formula (3)}\\x^6 x^{-12}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $x^6 x^{-12}$ .

So, the answers can be:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

8 0

3 years ago

A movie theater has been charging $10.00 per person and selling about 500 tickets on a typical weeknight. After surveying their

kifflom [539]

Answer:

$p(x) = -0.01x + 15$

Step-by-step explanation:

Given

Let x represent the number of tickets, and p the charges

$(x_1,p_1) = (500,10)$

A reduction of 50c gives an increment of 50 tickets.

This gives:

$(x_2,p_2) = (500+50,10-0,5)$ ---- $50c = \$0.5$

$(x_2,p_2) = (550,9.5)$

Required

Determine the demand function

First, calculate the slope:

$m = \frac{p_2 - p_1}{x_2 - x_1}$

$m = \frac{9.5 - 10}{550-500}$

$m = \frac{-0.5}{50}$

$m = -0.01$

So, the equation is:

$p = m(x - x_1) + y_1$

$p = -0.01(x - 500) +10$

$p = -0.01x + 5 +10$

$p = -0.01x + 15$

Hence, the function is:

$p(x) = -0.01x + 15$

4 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago