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3 years ago

Jane had 5 candy canes Max had 11 candy canes how many candy canes do they have all together

1 answer:

6 0

To answer this question, simply add together the amount of candy canes Jane has with the amount of candy canes Max has!

When Max = 11 and Jane = 5 ....

Max + Jane = 16

Hope this helps! :)

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Pleaseee help! No files please. thanks! ASAP!

Ainat [17]

Answer:

Step-by-step explanation:

use a protractor to mmeasure it

5 0

2 years ago

Please help me, i have been on this problem for 45 minutes.

docker41 [41]

Answer:

Step-by-step explanation:

The absolute difference is 77-(-8)=85. The only option that is equivalent is

|-8|+77

8 0

3 years ago

Solve and graph 6t+-3 greater then -2 t-19

Dafna11 [192]

Answer:

t > 4

Step-by-step explanation:

Rearrange:

Rearrange the equation by subtracting what is to the right of the greater than sign from both sides of the inequality :

-6*t-3-(-2*t-19)<0

Step by step solution :

STEP 1

: Pulling out like terms

Pull out like factors :

-4t + 16 = -4 • (t - 4)

Equation at the end of step 1:

STEP 2:

Divide both sides by -4

Remember to flip the inequality sign:

Solve Basic Inequality :

Add 4 to both sides

T > 4

Inequality Plot :

inequality plot for

-4.000 X + 16.000 > 0

( SEE ATTATCHMENT)

One solution was found :

t > 4

3 0

3 years ago

Please verify the following trigonometric identities.

Nina [5.8K]

Seems like there is a correction in the first question (RHS is tanx.tany)

(i) For convenience: let tanx = a ; tany = b

Thus, tanx + tany = a + b

Moreover, cotx = 1/tanx = 1/a ; coty = 1/b

Thus,

cotx + coty = 1/a + 1/b = (a + b)/ab

Hence,

=> (tanx + tany)/(cotx + coty)

=> (a + b) / { (a + b)/ab }

=> ab(a + b)/(a + b)

=> ab => tanx.tany , proved.

(ii) For convenience: let sinx = a ; cosx = b

As we know, sin²x + cos²x = 1 => a²+b²=1

=> (a³ + b³)/(a + b)

=> (a + b)(a² + b² - ab) / (a + b)

=> (a² + b² - ab)

=> 1 - ab => 1 - sinx.cosx , proved

(iii): let x/2 = A

=> tan(x/2) + cosx.tan(x/2)

=> tanA + cos2A.tanA

=> tanA [1 + cos2A]

=> tanA (2cos²A) {1+cos2A = 2cos²A}

=> (sinA/cosA) (2cos²A)

=> sinA (2cosA)

=> 2sinAcosA

=> sin2A

=> sin2(x/2)

=> sinx proved

Letting x/2 = A is not mandatory. I did it to decease words*(in a line).

Indentities used:

• sin²A + cos²A = 1

• (a³ + b³) = (a + b)(a² + b² - 1)

• 1 + cosA = 2 cos²(A/2)

• tanA = sinA/cosA.

• 2sinAcosA = sin2A

8 0

3 years ago

Read 2 more answers

Jack has some marbles. He has ten more red marbles than blue marbles. He has fewer yellow marbles than blue marbles.

kap26 [50]

Answer:

Jack has at least 15 marbles: 12 red, 2 blue and 1 yellow.

Step-by-step explanation:

1. Let's review the information given to us to solve the problem correctly:

Number of red marbles = 10 more than blue marbles

Number of yellow marbles fewer than blue marbles

2. How many of each color marble could Jack have?

r = Number of red marbles

b = Number of blue marbles

y = Number of yellow marbles

r = b + 10

y < b

If y = 1, then b ≥ 2, and r = b + 10,

fulfilling the three conditions, we have:

y = 1, b = 2, and r = 12

5 0

3 years ago