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svetoff [14.1K]
2 years ago
10

How do you this .. I don’t know how to do it

Mathematics
1 answer:
IRISSAK [1]2 years ago
7 0
I would help but your missing parts of the problem
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Which of the following statements are true of the graph of the function f(x)=(x+5)(x-3)?
musickatia [10]
You can use a calculator for this if you have the graphing kind (I use a Ti-84) and when plugging the equation in, the answers should be b, d, and e 
7 0
3 years ago
Yaaay even more algebra
Anika [276]

Answer:

f(x) = 3x-1

g(x) = x+2

Find (f-g)(x), this means we need to subtract g from x.

(3x-1) - (x+2)

Subtract the like terms:

3x -x = 2x

-1 -2 = -3

Combine them to get 2x-3

The answer is B.

Brainliest would be much appreciated!

5 0
3 years ago
At the market, 2.4 pounds of carrots cost $1.68 and 1.8 pounds of broccoli cost $1.53. How much more does 1 pound of broccoli co
Greeley [361]

For calculating that, you have to find the unit rate first;

2.4 pound carrot = $1.68

1 pound carrot = $1.68/2.4 = $0.7

1.8 pound broccoli = $1.53

1 pound broccoli = $1.53/1.8 = $0.85

So, it would be = $0.85 - $0.7 = $0.15

SO, YOUR ANSWER IS $0.15

5 0
4 years ago
Helppppppppppppppppppp I’ll mark you brainlist don’t respond if you’re going to put something random I will report you :)
Basile [38]

Answer:

35°

Step-by-step explanation:

Angles on a straight line add up to 180, so 180-145 is equal to

3 0
3 years ago
Please someone help me to prove this..​
Pachacha [2.7K]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}

\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}

\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}

\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}

\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}

\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10

LHS = RHS:  cot 10 = cot 10   \checkmark

8 0
3 years ago
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