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2 years ago

Helpppppppppp plssssssssss

Mathematics

1 answer:

Serhud [2]2 years ago

3 0

Answer:

12 in.

Step-by-step explanation:

We already know that the bottom line all together is 19 inches, and the line for the top of the square box at the end is also 7in. We can then infer that the top line of 7 cm is the same as the bottom line, so we therefore do 19 - 7 which equals to 12. Hope this helps.

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Consider the graph of the linear function h(x) = –6 + x. Which quadrant will the graph not go through and why?

Lesechka [4]

H ( x ) = - 6 + x
m = 1 ( the slope )
b = - 6 ( y - intercept )
x - intercept:
0 = - 6 + x
x = 6
The graph is going through Quadrants: I, II and IV.
Answer:
B ) Quadrant II, because the slope is positive and y-intercept is negative.

6 0

3 years ago

Read 2 more answers

GEOMETRY PROBLEM THAT I CAN NEVER GET

alina1380 [7]

Answer:

Therefore the friends will have to buy 511 feet of fence (rounding up) or 510 feet of fence ( to the nearest foot).

Step-by-step explanation:

i) perimeter of rectangular yard = 319 feet

ii) perimeter of yard of friends $= \frac{8}{5} \times 319 = 510.4$

iii) Therefore the friends will have to buy 511 feet of fence or 510 feet of fence ( to the nearest foot).

6 0

3 years ago

This is one way to subtract 8.98 from 14.32. What is another correct way?

Oxana [17]

Answer:

Here are a few:

14.32 - 8.98
-8.98 + 14.32

14.32

Hopefully this made sense and is the correct way to solve your problem! :)

5 0

3 years ago

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

3 years ago

Can someone help me with 3,4 and 5? For 3 and 4 it's asking "what is the volume of each solid? Round to the nearest tenth if nec

Marianna [84]

Answer:

the answer 3,b

4,f

5,c

7 0

3 years ago