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valentina_108 [34]
3 years ago
11

What is 76.78 rounded to the nearest whole number?

Mathematics
2 answers:
cricket20 [7]3 years ago
7 0
Your answer would be 77. 78 is closer to 100 (77) than it is to 0 (76).
BartSMP [9]3 years ago
3 0
The answer is 77 because it is only .22 away from 77 and .78 from 76 hoped this helped 
Feel free to ask anymore questions here ta brainly.com
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what is the inverse of h(x)=12e^2x. based on your answer write an expression write an expression equivalent to where x is h of 1
dezoksy [38]
Let y = 12 e^2x

e^2x = y/12
Taking [email protected]
ln e^2x = ln (y/12)

2x = ln (y/12)

x = (1/2) ln (y / 12)

so the inverse h-1(x)  = (1/2) ln ( x / 12)


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3 years ago
What is the area of a square with the side length 3a
ladessa [460]

the area of a square = length*length =3a*3a=9a^2

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What is the quotient of (x3 + 6x2 + 11x + 6) ÷ (x2 + 4x + 3)? x + 2 x – 2 x + 10 x + 6
evablogger [386]

Answer:

x + 2

Step-by-step explanation:

If we know one of the factors, we should be able to use long division.

We can divide the trinomial by the binomial and find the other factor.

                 <u>x  +  2</u><u>                   </u>

x² + 4x + 3)x³ + 6x² + 11x + 6

                  <u>x³ + 4x² + 3x</u>

                         2x² + 8x + 6

                         <u>2x² + 8x + 6</u>

                                           0

Thus,

\dfrac{x^{3} + 6x^2 + 11x + 6} {x^{2} + 4x + 3} = \mathbf{x + 2}

6 0
3 years ago
*will give brainliest*
emmasim [6.3K]

Answer:

Y = 6 and x=6 are perpendicular because they both measure at a 90 degree angle

7 0
3 years ago
Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
I am Lyosha [343]

Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
3 years ago
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