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viktelen [127]
2 years ago
11

Need help with question?

Mathematics
2 answers:
Nikolay [14]2 years ago
5 0

Answer:

150

Step-by-step explanation:

15 times 8 = 120

14-8 = 6

6 times 5 = 30

150

professor190 [17]2 years ago
5 0
Ooh yes I love answering these

First, you have to find the area
15 x 8 = 120
6 x 5 = 30
Add : 150
So the answer is A
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Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots StartRoo
anastassius [24]

Answer:

a) f(x)=3x^{3}-6x^{2}-15x+30

Step-by-step explanation:

1) In this question we've been given "a", the leading coefficient. and two roots:

x_{1}=\sqrt{5}\:x_{2}=2

2) There's a theorem, called the Irrational Theorem Root that states:

If one root is in this form x'=\sqrt{a}+b  then its conjugate x''=\sqrt{a}-b. is also a root of this polynomial.

Therefore

x_3=-\sqrt{5}

3) So, applying this Theorem we can rewrite the equation, by factoring. Remembering that x is the root. Since the question wants it in this expanded form then:

f(x)=3(x-\sqrt{5})(x+\sqrt{5})(x-2)\Rightarrow 3x^{3}-6x^{2}-15x+30

4 0
3 years ago
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What is the awnser.i don't get it.
Dennis_Churaev [7]
8x+Y=-16
answer
Y=-8x-16
6 0
3 years ago
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Which equations have the same value of x as Three-fifths (30 x minus 15) = 72? Select three options.
DanielleElmas [232]

Answer:

C 18 x minus 9 = 72

D 3 (6 x minus 3) = 72

E   x = 4.5

Step-by-step explanation:

Hope this helped

6 0
3 years ago
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What is the function written in vertex form?
lys-0071 [83]

Answer:

The answer in the procedure

Step-by-step explanation:

The question does not present the graph, however it can be answered to help the student solve similar problems.

we know that

The equation of a vertical parabola into vertex form is equal to

f(x)=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex

If the coefficient a is positive then the parabola open up and the vertex is a minimum

If the coefficient a is negative then the parabola open down and the vertex is a maximum

case A) we have

f(x)=3(x+4)^{2}-6

The vertex is the point (-4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case B) we have

f(x)=3(x+4)^{2}-38

The vertex is the point (-4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

case C) we have

f(x)=3(x-4)^{2}-6

The vertex is the point (4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case D) we have

f(x)=3(x-4)^{2}-38

The vertex is the point (4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

4 0
3 years ago
(t,-3) and (2,6); Slope = -1
tangare [24]
-1=\frac{Δy}{Δx}
-1=\frac{6-(-3)}{2-t}
t-2=9
t=11
4 0
3 years ago
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