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3 years ago

Can someone help me real quick

Mathematics

1 answer:

nata0808 [166]3 years ago

7 0

Answer: 7 rides

Step-by-step explanation:

Since it takes $4 to enter the fair, you have

$21.50 - $4 = $17.50 left to spend on rides

$17.50/$2.50 = 7 rides

hope this helps!

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PLEASEEEEEE SOMEONE HELP I WILL GIVE ALOT OF POINTS AND WHATVER ELSE PLEASEE

Triss [41]

Part A : formula for circumference = pi*D or 2piR
for circle A.... 8*pi=25.12
therefore pi=3.14
for circle B.... 3*pi=9.42
therefore pi=9.42/3
pi=3.14

part B : formula for Area =
$pi \times {r}^{2}$
for circle A.... 50.24=pi*(8/2)sqr...(because diameter/2= radius)
pi=50.24/16
pi=3.14

for circle B... 7.065=pi*(3/2)sqr...
pi=7.065/2.25
pi=3.14

what do you observe?
the values for pi are all equal... therefore pi is constant.

4 0

4 years ago

Ned and Jack are each saving money for a new smartphone. Ned initially has $65 and plans to save $5.50 per week. Jack does not h

julia-pushkina [17]

Answer:

I'm not sure exactly what you're looking but I would set up the equation like this.

y=65+5.5x for Ned's equation

y=8.75x for Jack's equation.

Step-by-step explanation:

This is in y=mx+b format, hopefully it helps!

7 0

3 years ago

After saving 53% of his allowance, spending 26% of his allowance on food, and using $2 of the $20 allowance to pay library fines

Flura [38]

His allowance is $20
He saves 20(.53)
He spends 20(.26)
He uses $2

20 - 20(.53) - 20(.26) - 2 = Money left

6 0

3 years ago

Read 2 more answers

Solve the equation after making an appropriate substitution. X^6-98x^3=3375

Leni [432]

Rearrange $x^6-98x^3=3375$ to give

$x^6-98x^3-3375=0$

Let $x^3$ be a variable 'p' and so we can write $x^6$ as $x^6=(x^3)(x^3)=(p)(p)= p^2$

Rewrite the equation in terms of 'p'

$p^2-98p-3375 = 0$

where $a=1, b=-98, c=-3375$

using the quadratic formula $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ and subsitute the value of $a, b, c$

$p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125$

$p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27$

There are two value of p; 125 and -27

Now we find the value of x

Earlier we substitute $x^3$ for $p$ and $x^6$ for $p^2$

When $p=125$ , $x= \sqrt[3]{125}=5$
When $p = -27, x= \sqrt[3]{-27}=-3$

So the final answer is the two values of x;

x = 5 OR x = -3

3 0

3 years ago

A. 0 B. 1 C. 2 D. 3

zloy xaker [14]

-19 to -15 would be a positive 4 difference

19 to 15 = 4 this would be a -4 difference

3 to 7 is a positive 4 difference

-3 to 7 is a difference of 10

so there is 1 with a -4 difference
answer is B.

6 0

3 years ago