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3 years ago

Please help with the first one the 5d

Mathematics

1 answer:

qwelly [4]3 years ago

7 0

It would be 4d because there is a one in from the d that is by itself but you cannot see it and therefore 5d-1d= 4d.

Hope this helps! :)

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Which of the following is a factor of 3x3-12x2 +X-4 ?<br> A. X+3<br> B. X+2<br> C. X-4<br> D.X-1

DiKsa [7]

Answer:

B.x+2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Use the stem and leaf plot graph to answer the question.

kogti [31]

<h3>Answer: 13 days</h3>

====================================================

Explanation:

Count the number of leaves marked in green.

Row One: 2 leaves (1 and 8)
Row Two: 3 leaves (3, 6, and 9)
Row Three: 5 leaves (2, 3, 4, 6, 7)
Row Four: 3 leaves (1, 6, and 8)

There are 2+3+5+3 = 13 leaves total. This represents tickets were sold on 13 days.

Each stem and leaf combines to form a different ticket count. For instance, the stem of 1 and leaf of 8 combine to get the value 18. So 18 tickets were sold on that particular day.

7 0

1 year ago

Write (-5+8i)+(-5-8i) as a complex number in standard form

gizmo_the_mogwai [7]

$(-5+8i)+(-5-8i) =
-5+8i-5-8i=-10$

5 0

3 years ago

What is mn+5m-2+mn+3

viktelen [127]

Answer:

$mn + 5m - 2 + mn + 3 \\ = 2mn + 5m + 1 \\ = = = = = = = = = = =$

<h2>HOPE U UNDERSTOOD</h2>

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4 0

3 years ago

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

<u>Hence final answer is -0.47.</u>

7 0

3 years ago