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3 years ago

I really need help on all of them, I will give out brainliest to one answer and ten points to both

Mathematics

1 answer:

tangare [24]3 years ago

3 0

Answer:

1. 1/2, 2. 2, 3. 1/-3, 4. -2

Step-by-step explanation:

1) I know that the equation for slope is y2-y1/x2-x1. For number one I first found two points of the graph which is (0,3) and (2,4) (not sure if I see it correctly, its kind of blurry). I decided that (0,3) would be my y1 and x1 and (2,4) will be my y2 and x2. I later plugged the number in the equation and got 4-3/2-0 which the answer turns out to be 1/2.

2) For the rest of the questions it is kinda the same business as question 1. For question two I chose two points out of the table which wrer (-1, -12) and (1,-8) and plugged them into the equation and got -8-(-12)/1-(-1) which the answer turns out to be 2.

3) similar to question 2. The two points(2,-6) and (-4,-3) are given to you. so I plugged them in again and got -3-(-6)/-4-2 which the answer turns out to be 3/-6 into 1/-3.

4) Same with question 1 find two points of the graph. I chose (0,3) and (1,1) (again might not be accurate it's kind of blurry) I plugged them into the equation again and got 1-3/1-0 which equals -2.

Just always remember y2-y1/x2-x1=m is the equation for slope.

(m is the variable for slope)

Hoped this helped you : )

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Write the domain as an inequality <br> &<br> Write the range as an inequality

vesna_86 [32]

Answer:

D = x < — 4

R = y < — 3

Step-by-step explanation:

If any questions remains, ask in the comment <3

7 0

2 years ago

Please help me with these, oh sweet jesus

Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

80. $\sin^{4}x - \cos^{4}x = 1 - 2\cos^{2}x + 2 \cos^{4} x$ Proved.

Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

= $1 - 2\cos^{2}x + 2 \cos^{4} x$

= Right hand side. (Proved)

7 0

3 years ago

A math class has 3 girls and 7 boys in the 7th grade and 5 girls and 5 boys in the 8th grade the teacher randomly selects a seve

Minchanka [31]

In the 7th and 8th grade combined, there 3+5 girls = 8 girls, and 7+5 boys = 12 boys. If there are only boys and girls, then there are 12+8=20 students in all.

The fraction of girls out of the total is P = 8/20 = 2/5.

8 0

3 years ago

GRAPHING INEQUALITIES

Rudiy27

I don’t know put 2x on the x axis and dot 0 on the y axis

6 0

3 years ago

can someone answer these 2 questions asap and thank you its also double the points because it 2 questions ;)

Juli2301 [7.4K]

r $\leq$ 15.5 (not the one with the line under it i just couldnt find the one that didnt have the line under it)

for the graph, put 16 on the right side and label the lines right to left 16 15.5 14 14.5 etc. Put an open circle on the 15.5 line. then make an arrow pointing to the left.

10:

80 $\leq$ p+64

This is because she needs 80 points OR MORE and (p represents the points) she already has 64. She needs 80 to be less than or equal to the amount of points earned.

The solution is 16 $\leq$ p because 80-64=16 and p is needs to be greater than or equal to 16

7 0

3 years ago