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3 years ago

Can someone please assist me on this problem, any help would be appreciated.

Mathematics

1 answer:

sashaice [31]3 years ago

3 0

Answer:

11x^2 -4x + 3

Step-by-step explanation:

(x^2-5x+7)+(10x^2+x-4)

= x^2-5x+7+10x^2+x-4

= 11x^2 -4x + 3

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A box is filled with 3 yellow cards , 4 green cards. A card is chosen at random from the box. What is the probability that the c

nasty-shy [4]

Answer:

3/7

Step-by-step explanation:

total cards 3+4 = 7

Not green = 7-4 = 3 cards

P ( not green )= not green cards / total = 3/7

6 0

3 years ago

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Explain whether each equation is a linear equation.

Aleksandr-060686 [28]

Answer:

3. Not linear

4. Linear

Step-by-step explanation:

3. The equation has an exponent on the x. Since x is squared the graph is a parabola. The graph is a curve, not a line, so it is not linear.

4. Both the x and y have no exponent (which is an exponent of 1) The graph will be a line. This is a linear function.

8 0

2 years ago

Someone please help I want to double check PLEASEEEEE HELLPPPP

nikitadnepr [17]

Answer:

correct

Step-by-step explanation:

6 0

3 years ago

Rewrite the standard form equation below into slope-intercept form.<br>x – 2y = 3

Gnom [1K]

X-2y = 3

-2y=3-x move the x over to the other side

-y=3/2 -1/2x divide both sides by 2

y=1/2x - 3/2 divide both sides by -1

8 0

3 years ago

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Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations

marusya05 [52]

Answer:

$Sin \angle A =0.80$

$Cos \angle A=0.60$

$Sin \angle B =0.60$

$Cos \angle B=0.80$

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle A =\frac{BC}{BA}$

Substitute values for BC and BA

$Sin \angle A =\frac{8cm}{10cm}$

$Sin \angle A =\frac{8}{10}$

$Sin \angle A =0.80$

$Cos \angle A=\frac{AC}{BA}$

Substitute values for AC and BA

$Cos \angle A=\frac{6cm}{10cm}$

$Cos \angle A=\frac{6}{10}$

$Cos \angle A=0.60$

Solving (b): Sine and Cosine B

In trigonometry:

$Sin \theta =\frac{Opposite}{Hypotenuse}$ and

$Cos \theta =\frac{Adjacent}{Hypotenuse}$

So:

$Sin \angle B =\frac{AC}{BA}$

Substitute values for AC and BA

$Sin \angle B =\frac{6cm}{10cm}$

$Sin \angle B =\frac{6}{10}$

$Sin \angle B =0.60$

$Cos \angle B=\frac{BC}{BA}$

Substitute values for BC and BA

$Cos \angle B=\frac{8cm}{10cm}$

$Cos \angle B=\frac{8}{10}$

$Cos \angle B=0.80$

Using a calculator:

$A = 53^{\circ}$

So:

$Sin(53^{\circ}) =0.7986$

$Sin(53^{\circ}) =0.80$ -- approximated

$Cos(53^{\circ}) = 0.6018$

$Cos(53^{\circ}) = 0.60$ -- approximated

$B = 37^{\circ}$

So:

$Sin(37^{\circ}) = 0.6018$

$Sin(37^{\circ}) = 0.60$ --- approximated

$Cos(37^{\circ}) = 0.7986$

$Cos(37^{\circ}) = 0.80$ --- approximated

8 0

3 years ago

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