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mestny [16]
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
Mathematics
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
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