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slavikrds [6]
3 years ago
6

Answer to question 5

Mathematics
1 answer:
kogti [31]3 years ago
7 0
Three plus parenthesis four times twelve
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If h(x) = 5x − 3 and j(x) = −2x, solve h[j(2)] and select the correct answer below.
Sati [7]
Sub 2 for x in j(x) and evalueate
then sub that result for x in h(x)

j(2)=-2(2)
j(2)=-4

now sub-4 for  x in h(x)

h(-4)=5(-4)-3
h(-4)=-20-3
h(-4)=-23

answer is -23
8 0
3 years ago
I need the answer for #13
ira [324]
Is that the top or bottom? Assuming it's the top, when distributing the -2 to what's inside the parenthesis, they were correct in multiplying -2•7, but when they multiplied -2 and -y, the answer needs to be positive 2y because they are multiplying double negatives
8 0
4 years ago
Read 2 more answers
Y = x² + 6x + 3
olganol [36]

Vertex<em> </em>is at \left(-3,-6\right)

<em>y-intercept</em> is 3.

The parabola <em>opens up</em>.

Step-by-step explanation:

The graph of the equation is hereby attached in the answer area.

Vertex is the point on the parabola where the graph crosses its axis of symmetry. The axis of symmetry here(x =-3), is shown with the dotted line in the graph attached.

<em>y-intercept </em>is defined as the value of y where the graph crosses the y-axis. In other words, when x=0. Putting

And, the graph opens up as shown the graph figure as well. It is also evident from the co-efficient of x in the given equation y =x^{2} + 6x + 3. Here, co-efficient of

So, vertex<em> </em>is at \left(-3,-6\right)

<em>y-intercept</em> is 3.

The parabola <em>opens up</em>.

7 0
3 years ago
Find the number of terms in this polynomial. -5h + 8​
Arisa [49]

Hi! I can help you with joy! :)

  • Let's find the number of terms in this polynomial.
  • The given polynomial has 2 terms:
  • -5h and 8
  • Polynomials with 2 terms are called "binomials"
  • Thus, -5h+8 has 2 terms.

<u>Answer:</u>

2 terms

Hope it helps!

\rm{Stargazing{\bold{WithJoy:D}

4 0
3 years ago
Someone please help me!<br><br> 64x squared y cubed + 8xy divided by 4xy squared
Burka [1]
64x^2+8xy/4xy^2. This should be what you’re looking for hopefully
7 0
3 years ago
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