<img src="https://tex.z-dn.net/?f=5%3A8%3Dn%3A24" id="TexFormula1" title="5:8=n:24" alt="5:8=n:24" align="absmiddle" class="late

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ohaa [14]

3 years ago

x-formula"> $\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

Mathematics

1 answer:

klemol [59]3 years ago

6 0

QUESTION:

$5:8=n:24$ $\left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]$

$\frac{5}{8} = \frac{n}{24} \\ 5 \times 24 = 120 \\ 8 \times n = 8n \\ 120 \\ 8n \\ = 15$

ANSWER:

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Mary used Delicious (d) and Golden Delicious (g) apples to make homemade applesauce. Delicious apples are $0.75 each and Golden

alina1380 [7]

Answer:

Mary bought 9 Golden Delicious apples

Step-by-step explanation:

This can be solved by setting up a system of equations. Based on the information given, the equations would be set up as follows.

$.75d+1.25g=21.00\\d+g=22$

I prefer to solve using substitution, but elimination can be used as well

$.75d+1.25g=21.00\\g=22-d$

We can now plug in this g value to the first equation

$.75d + 1.25 (22-d)=21.00\\.75d + 27.5-1.25d=21\\27.5-0.5d=21\\-.5d=-6.5\\d=13$

This means mary bought 13 delicious apples. She bought 22 apples in total, so the total number minus the amount of delicious apples will give us the amount of golden delicious apples.

22 - 13 = g

9 = g

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<em>Let me know if you have any questions! Spread the love. </em>

3 0

3 years ago

What is a counterexample for "subtraction of whole numbers is commutative" ??

devlian [24]

The commutatice property for addition says a+b=b+a
If it was for subtraction, then it would be a-b=b-a
Take a=7 and b=5
7-5 is not equal to 5-7
2 is not equal to -2
So subtraction is not commutative for whole numbers.

Hope this helps :)

6 0

4 years ago

10 weeks ago Jerry bought stock at 21 and a half today the stock is valued at 20 and 3/8 we can say the stock is performing at w

seropon [69]

Par value means that the stock is quoted at the face or principal value.

If Jerry bought the stock at the face value, it means that the stock is now below its face value, and that is said that the stock is below par.

Therefore, the answer is below par.

Answer: below par.

7 0

3 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago

P(x)=(x-2)(x-3)(x+7) find the constant term

hram777 [196]

Short method: multiply constants
$= ( - 2) \times ( - 3) \times (7) \\ = 6 \times 7 \\ = 42$

Long mothod: expand it
$= ( {x}^{2} - 5x + 6)(x + 7) \\ = ... \\ = ... + 42$

Answer: 42

5 0

3 years ago