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2 years ago

Can somebody HELP me as soon as possible.

Mathematics

2 answers:

Butoxors [25]2 years ago

4 0

Answer:

A: Linear

B: Nonlinear

C: Linear

solmaris [256]2 years ago

3 0

Answer:

A.liner

B.non liner

C.liner

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Can someone solve this for me pls

adelina 88 [10]

Answer:

a= m<SQR

Step-by-step explanation:

8 0

2 years ago

a person Can travel 120 miles in one direction. the return trip was accomplished at double the speed and took three hours less t

BaLLatris [955]

40 mph would be 120 miles in 3 hours (120/40)

80 mph is 120 miles in 1.5 hours(120/80)

3 - 1.5 = 1.5, not 3 hrs

Change the speed to 20 mph( 1/2 the speed) to double the hours.

20 mph = 6 hours(120/20)

40 mph = 3 hrs(120/40)

6 - 3 = 3 hours difference

6 0

3 years ago

M is the midpoint of CD. Given the coordinates of one endpoint C(10,-5) and the midpoint M(4, 0), find the coordinates of the ot

mash [69]

Answer:

D(-2, 5).

Step-by-step explanation:

We are given that M is the midpoint of CD and that C = (10, -5) and M = (4, 0).

And we want to determine the coordinates of D.

Recall that the midpoint is given by:

$\displaystyle M = \left(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2}\right)$

Let C(10, -5) be (x₁, y₁) and Point D be (x₂, y₂). The midpoint M is (4, 0). Hence:

$\displaystyle (4, 0) = \left(\frac{10+x_2}{2} , \frac{-5+y_2}{2}\right)$

This yields two equations:

$\displaystyle \frac{x_2 + 10}{2} = 4\text{ and } \frac{y_2 - 5}{2} = 0$

Solve for each:

$\displaystyle \begin{aligned}x_2 + 10 &= 8 \\ x_2 &= -2 \end{aligned}$

And:

$\displaystyle \begin{aligned} y_2 -5 &= 0 \\ y_2 &= 5\end{aligned}$

In conclusion, Point D = (-2, 5).

5 0

2 years ago

Diana has just begun the process of filing her federal income tax return, and she plans to deduct medical and dental expenses. w

Andru [333]

Answer:

1040 tehyimngtumh

Step-by-step explanation:

Apex

5 0

3 years ago

PLEASE ANSWER + BRAINLIEST!! Factor completely. 4k - 20k^9 = 3b^2 - 108 =

Neporo4naja [7]

$4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)$

$3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)$

$Used:\ (a-b)(a+b)=a^2-b^2$

7 0

3 years ago