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Andreas93 [3]
3 years ago
7

Pls help me with these math problems

Mathematics
2 answers:
Fittoniya [83]3 years ago
4 0
Well for 2 you have to turn the subtraction sign to addition
user100 [1]3 years ago
3 0
2. W<5
4. D<6 (or equal to)

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A standard container measures 2.4 m by 2.4 m by 6 m.
ella [17]

Answer:

34.56

Step-by-step explanation:

5 0
3 years ago
HELP ASAP<br> 6+ x - y + 10=
Katena32 [7]

Answer:

x-y+16

Step-by-step explanation:

add the like terms, 6 and 10 which is 16. since the variables are not like terms, thats your final answer. hope it helped :)

7 0
3 years ago
Draw to explain how you would make a ten to find 5+8.
erica [24]
Since 2+8 =10 you would just have 3 left to add so it would be 10+3=13
4 0
3 years ago
Which term could be put in the blank to create a fully simplified polynomial written in standard form?
Katena32 [7]

Correct Question:

Which term could be put in the blank to create a fully simplified polynomial written in standard form?

8x^3y^2 -\ [\ \ ] + 3xy^2 - 4y3

Options

x^2y^2            x^3y^3         7xy^2         7x^0y^3

Answer:

x^2y^2

Step-by-step explanation:

Given

8x^3y^2 -\ [\ \ ] + 3xy^2 - 4y^3

Required

Fill in the missing gap

We have that:

8x^3y^2 -\ [\ \ ] + 3xy^2 - 4y^3

From the polynomial, we can see that the power of x starts from 3 and stops at 0 while the power of y is constant.

Hence, the variable of the polynomial is x

This implies that the power of x decreases by 1 in each term.

The  missing gap has to its left, a term with x to the power of 3 and to its right, a term with x to the power of 1.

This implies that the blank will be filled with a term that has its power of x to be 2

From the list of given options, only x^2y^2 can be used to complete the polynomial.

Hence, the complete polynomial is:

8x^3y^2 -x^2y^2+ 3xy^2 - 4y3

4 0
3 years ago
Read 2 more answers
-25 Points-
Over [174]

\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{s}{s}=\cfrac{\sqrt[3]{8}}{\sqrt[3]{125}}\implies \cfrac{s}{s}=\cfrac{2}{5}\qquad \leftarrow \textit{ratio of the sides or scale factor}

8 0
3 years ago
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