Ask question

Ask question

All categories

3 years ago

6

Rowan has 50$ in a saving jar and is putting in 5$ every week. Jonah has 10$ in his own jar is putting 15$ very week. Each of th

em plots his progress on a graph with time on horizontal axis and amount in the jar on the. Vertical axis. Which statement about their graph is true.
(A) Rowan's graph has a steeper slope than Johan's
(B) Rowan's graph always lies above Johan's
(C) Johan graph had a steeper slope than rowan
(D) Johan graph always lies above rowan

1 answer:

Misha Larkins [42]3 years ago

3 0

C is true because the slope for johan is 15 compared to Rowans slope which is 5

You might be interested in

How to reduce into simpler terms.?

Sveta_85 [38]

Answer:

The simplest form of the fraction $\frac{45}{100}$ is $\frac{9}{20}$ .

i.e.

$\frac{45}{100}=\frac{9}{20}$

Step-by-step explanation:

Here are some simple observations regarding how to reduce a fraction into simpler terms:

A fraction is reduced to lowest or simplest terms by finding an equivalent fraction in which the numerator and denominator are as small as possible.

In order to reduce a fraction to lowest or simplest terms, divide the numerator and denominator by their (GCF). Note that (GCF) is also called Greatest Common Factor .

So, lets take a sample fraction and reduce into simpler terms.

Considering the fraction

$\frac{45}{100}$

$\mathrm{Find\:a\:common\:factor\:of\:}45\mathrm{\:and\:}100\mathrm{\:in\:order\:to\:cancel\:it\:out}$

$\mathrm{Greatest\:Common\:Divisor\:of\:}45,\:100:\quad 5$

$\mathrm{Factor\:out\:}5\mathrm{\:from\:the\:numerator\:and\:the\:denominator}$

$45=5\cdot \:9\mathrm{,\:\quad }100=5\cdot \:20$

so

$\frac{45}{100}=\frac{5\cdot \:\:9}{5\cdot \:\:20}$

$\mathrm{Cancel\:the\:common\:factor:}\:5$

$=\frac{9}{20}$

Therefore, the simplest form of the fraction $\frac{45}{100}$ is $\frac{9}{20}$ .

i.e.

$\frac{45}{100}=\frac{9}{20}$

4 0

3 years ago

Which equation represents a nonproptional relationship

Luba_88 [7]

I need to see a picture

7 0

3 years ago

Simplify the expression. 35(12)(5)

Furkat [3]

Answer:

the answer is 2100

Step-by-step explanation:

7 0

3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

4 0

2 years ago

Find the value of the variable if P is between J & K. A) JP = 8z - 17 B) PK = 5z + 37 C) JK = 17z - 4

Daniel [21]

<span>We cannot deduce about the exact location of P between J and K. But we can conclude: segment JP + segment PK = line JK.
</span><span>JP + PK = JK.
</span><span>Substitute first each.
(8z - 17) + (5z + 37) = 17z - 4
Combine like terms.
13z + 20 = 17z - 4
Isolate the variable z.
4z = 24
z = 6
The value of the variable z is then 6 units.</span>

3 0

2 years ago

Read 2 more answers