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3 years ago

I need help please???!!!

Mathematics

1 answer:

mel-nik [20]3 years ago

6 0

Answer:

The order pair (4,-5) is NOT a solution of the inequality

Step-by-step explanation:

Plug in to see if the order pair (4,-5) is a solution of the inequality

y - 8 < - 4x

-5 - 8 ? -4( 4)

- 13 > -16

So answer:

The order pair (4,-5) is NOT a solution of the inequality

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7(x-2)=28 Can somebody help me and show the steps?

olya-2409 [2.1K]

Since the 7 is in front of the parentheses you must multiply it by everything inside the parentheses. 7 times X is 7X and 7 times -2 is -14. Then you need to make the X be by itself on one side, to do that you have to add 14 to -14 to make it disappear, and also add 14 to the other side, to 28. Then you divide both sides by 7 to make X be by itself.

7(x-2)=28
7X-14=28
7x=42
X=6

6 0

2 years ago

(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28

Veronika [31]

$(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}$

4 0

3 years ago

Read 2 more answers

Each of six jars contains the same number of candies. Alice moves half of the candies from the first jar to the second jar. Then

tino4ka555 [31]

Answer:

The number of candies in the sixth jar is 42.

Step-by-step explanation:

Assume that there are x number of candies in each of the six jars.

⇒ After Alice moves half of the candies from the first jar to the second jar, the number of candies in the second jar is:

$\text{Number of candies in the 2nd jar}=x+\fracx}{2}=\frac{3}{2}x$

⇒ After Boris moves half of the candies from the second jar to the third jar, the number of candies in the third jar is:

$\text{Number of candies in the 3rd jar}=x+\frac{3x}{4}=\frac{7}{4}x$

⇒ After Clara moves half of the candies from the third jar to the fourth jar, the number of candies in the fourth jar is:

$\text{Number of candies in the 4th jar}=x+\frac{7x}{4}=\frac{15}{8}x$

⇒ After Dara moves half of the candies from the fourth jar to the fifth jar, the number of candies in the fifth jar is:

$\text{Number of candies in the 5th jar}=x+\frac{15x}{16}=\frac{31}{16}x$

⇒ After Ed moves half of the candies from the fifth jar to the sixth jar, the number of candies in the sixth jar is:

$\text{Number of candies in the 6th jar}=x+\frac{31x}{32}=\frac{63}{32}x$

Now, it is provided that at the end, 30 candies are in the fourth jar.

Compute the value of x as follows:

$\text{Number of candies in the 4th jar}=40\\\\\frac{15}{8}x=40\\\\x=\frac{40\times 8}{15}\\\\x=\frac{64}{3}$

Compute the number of candies in the sixth jar as follows:

$\text{Number of candies in the 6th jar}=\frac{63}{32}x\\$

$=\frac{63}{32}\times\frac{64}{3}\\\\=21\times2\\\\=42$

Thus, the number of candies in the sixth jar is 42.

4 0

3 years ago

-5/4-(-1/6) help please

tiny-mole [99]

A way to add fractions that always works is to multiply each numerator by the denominator of the other, then express the sum of products over the product of the denominators.
$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{d}+\dfrac{c}{d}\cdot\dfrac{b}{b}\\\\=\dfrac{ad+bc}{bd}$

Here, you have
$\dfrac{-5}{4}-\dfrac{-1}{6}=\dfrac{-5\cdot6-(-1)\cdot4}{4\cdot6}=\dfrac{-26}{24}\\\\=\dfrac{-13}{12}=-1\frac{1}{12}$

The sum is -1 1/12

3 0

3 years ago

Two opposite sides of the square are increased by 8 inches and the other two sides are decreased by 5 inches. the area of the ne

Iteru [2.4K]

Answer:

The length of a side of the square would be 8.

Step-by-step explanation:

7 0

2 years ago