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Bad White [126]
4 years ago
15

Oneta writes an algebraic expression with three terms. The y-term has a coefficient of -3, and the x-term has a coefficient of 1

. The expression does not have a constant term. Which expression could she have written?
Mathematics
2 answers:
Anuta_ua [19.1K]4 years ago
6 0

Answer:

1x - 3y + z

Step-by-step explanation:

Oneta has to write an algebraic expression with three terms without constant.

Given: The coefficient of x is 1 and the coefficient of y is -3.

so the possibilities are:

1x - 3y + z which is three term without constant.

1x - 3y - z which is three terms without constant.

Which represents three-dimensional figure.

Rina8888 [55]4 years ago
3 0
X - 3y + z

Another term would be the 3rd term which would be working with 3d planes instead of 2d planes.
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Simplify the expression 6C5.<br> a.6.<br> b. 24<br> C. 30<br> d. 720
Lana71 [14]
<h3>Answer:  A) 6</h3>

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

Explanation:

Plug n = 6 and r = 5 into the nCr combination formula

_n C _r = \frac{n!}{r!*(n-r)!}\\\\_6 C _5 = \frac{6!}{5!*(6-5)!}\\\\_6 C _5 = \frac{6!}{5!*1!}\\\\_6 C _5 = \frac{6*5*4*3*2*1}{5*4*3*2*1*1}\\\\_6 C _5 = \frac{720}{120}\\\\_6 C _5 = 6\\\\

Or you could use the shortcut

_n C _{n-1} = n\\\\

Yet another path you could take is to use Pascal's Triangle. Locate the row that starts with 1,6,... and then locate the second to last item. That value in the triangle is 6.

A real world interpretation is to consider having 6 people and you are selecting 5 of them to form a group where order doesn't matter. How many ways are there to do this? Well there are 6 such ways because there are 6 ways to leave someone out of the group.

8 0
2 years ago
Question 3(Multiple Choice Worth 3 points)
Marrrta [24]

Hello,

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8 0
3 years ago
Use the graph of f at the right to complete each of the parts (a) through (h)
Maru [420]

<u><em>Answers:</em></u>

a. The domain of f is ]-∞ , ∞[

b. The range of f is [-4 , ∞[

c. f(1) = 5

d. The vales of x for which f(x) is -3 are 7 and 9

e. The points at which the graph crosses the x-axis are (6,0) and (10,0)

f. The point where the graph of f crosses the y-axis is (0,5)

g. The values of x for which f(x) < 0 are ]6 , 10[

h. f(-7) is positive

<u><em>Explanation:</em></u>

<u>Part a:</u>

The domain of the function refers to all the possible x-values that can be used as an input for this function.

Taking a look at the graph, we can see that the graph extends endlessly from both ends of the x-axis. This means that all x-values can be used as a domain for the function. In other words, the domain of the function is all the real numbers.

<u>In interval notation, this is written as:</u>

Domain = ]-∞ , ∞[

<u>Part b:</u>

The range of the function refers to all the possible y-values that can be used as an output for the function.

From the graph, we can note that the function extends endlessly in the direction of the positive y-axis while it stops at a value of -4 in the direction of the negative y-axis.

This means that the range of the function starts from -4 (included) and extends to positive infinity.

<u>In interval notation, this is written as:</u>

R = [-4 , ∞[

<u>Part c: </u>

f(1) means that we are looking for the output (the value of y) for which the input (the value of x) is 1.

From the graph, we start by searching for x=1 (first square edge after the origin) and then move vertically till we intersect the graph.

Doing this, we will find that the value of y at x=1 is 5

<u>Therefore:</u>

f(1) = 5

<u>Part d:</u>

f(x) = -3 means that the output value (the value of the y) for the certain input (value of x) is -3

To get the value of x, we go y=-3 and move horizontally till we intersect the graph.

<u>Doing this, we will find that</u> the value of y is -3 at x = 7 and x = 9

<u>Part e:</u>

The points where the graph crosses the x-axis are the points that have y-value equal to 0

<u>Checking the graph, we can note that</u> the function crosses the x-axis at two points which are (6,0) and (10,0)

<u>Part f:</u>

The point where the graph crosses the y-axis is the point that has x-value equal to 0

<u>Checking the graph, we can note that</u> the function crosses the y-axis at only one point which is (0,5)

<u>Part g:</u>

f(x) < 0 means that the output of the function (the y-value) is less than 0 (0 is not included)

Taking a look at the graph, we can note that the function has negative output on the interval from 6 (excluded) to 10 (excluded)

<u>In interval notation, this is written as</u> ]6 , 10[

<u>Part h:</u>

f(-7) means the output of the function (the y-value) at input (x-vale) equal to -7

From the graph, we can note that the function has a constant value of 5 starting from x=5 till -∞

<u>This means that</u>, at x=-7, the value of y is 5 which is a positive value

Hope this helps :)

6 0
4 years ago
Read 2 more answers
In the function f(x) = 5(x2 − 4x + ____) + 15, what number belongs in the blank to complete the square?
Georgia [21]
If you would like to find the number which belongs in the blank to complete the square, you can calculate this using the following steps:

f(x) = 5(x^2 - 4x + a) + 15
a ... the number = ?

<span>x^2 - 4x + a = (x - 2)^2 = x^2 - 4x + 4
</span>a = 4

The correct result would be 4.
5 0
3 years ago
Read 2 more answers
Let U = {q, r, s, t, u, V, W, x, y, z}
Alexus [3.1K]

Answer:

B'=U-B={q, r, s, t, u, V, W, x, y, z}-{q, s, y, z}={r,t, u, V, W, x, y}

A'=U-A={q, r, s, t, u, V, W, x, y, z}-{q, s, u, w, y}={r,t,v,w,x,z}

B'UC={r,t, u, V,W, x, y}U{v, w, x, y, z).={{r,t, u, V,W, x, y,z}

Step-by-step explanation:

(B'U C) U A'={{r,t, u, V,W, x,y,z}U{r,t,v,w,x,z}={{r,t, u, V,W, x,y,z}

5 0
3 years ago
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