Juan purchased 16 lemons and used 5 on the first day of making lemonade. On the second day, he purchased 12 lemons and used 11. Juan wrote this expression to find how many lemons he should have left.

(16 – 5) + (12 – 11)

Answer:

x = -2±i

Step-by-step explanation:

Answer:

  -8c^4/m^3

Step-by-step explanation:

The applicable rules of exponents are ...

  (a^b)^c = a^(bc)

  a^-b = 1/a^b

__

  [tex](-2)^3\cdot(c^{-2})^{-2}\cdot m^{-6}=-8\cdot c^{(-2)(-2)}\cdot m^{-3}\\\\=\boxed{\dfrac{-8c^4}{m^3}}[/tex]

Answer:

It depends.

Step-by-step explanation:

If it is two straight lines, then no, there is no way for the lines to intersect in more than one place. If the line is a parabola, then the lines would intersect in 2 places. It depends on what type of line you have.  

Answer:

It is 4

Step-by-step explanation:

4

4 8

4 8 12

as you can tell 4 is your answer.

Answer:  9 1/10      91/10

Step-by-step explanation:

Answer:

9 1/10 and 91/10

Step-by-step explanation:

hope it helps :)

brainliest pls? have a good day/night

Answer: It is A

Step-by-step explanation:

In the given figure D) DC/BC has the same ratio as AB/BC.

The interior angle bisector theorem states that in a triangle the angle bisector of an angle divides the opposite side to the angle in the ratio of the remaining two sides.

In ΔABC m∠A = 28° and the ratio is taken for the sides which is opposite to the given angle BC and the side which not base AB.

Given the ratio AB/BC.

∴ In ΔBCD in which m∠B is also 28° we have to take the same sides opposite side is DC and the side which is not the base BC.

So, in the triangles ABC and ABD The ratio which has the same value as AB/BC is BC/DC.

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Hi there! :)

2t + 8 ≥ -4(t + 1)

Distribute the coefficient outside of the parenthesis:

2t + 8 ≥ -4(t) -4(1)

2t + 8 ≥ -4t - 4

Add 4t to both sides:

6t + 8 ≥ -4

Subtract 8 from both sides:

6t ≥ -12

Divide both sides by 6:

t ≥ -2.

Answer: t ≥  -2/3

Step-by-step explanation:

Answer:  see proof below

Step-by-step explanation:

Given: A + B + C = π              → A + B = π - C

                                              → B + C = π - A

                                              → C + A = π - B

                                              → C = π - (B +  C)

Use Sum to Product Identity:  cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B

Use the Double Angle Identity: sin 2A = 2 sin A · cos A

Use the Cofunction Identity: cos (π/2 - A) = sin A

Proof LHS → Middle:

[tex]\text{LHS:}\qquad \qquad \cos \bigg(\dfrac{A}{2}\bigg)+\cos \bigg(\dfrac{B}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)[/tex]

[tex]\text{Sum to Product:}\qquad 2\cos \bigg(\dfrac{\frac{A}{2}+\frac{B}{2}}{2}\bigg)\cdot \cos \bigg(\dfrac{\frac{A}{2}-\frac{B}{2}}{2}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{C}{2}\bigg)[/tex]

[tex]\text{Given:}\qquad \quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi -(A+B)}{2}\bigg)[/tex]

[tex]\text{Sum/Difference:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{2}\bigg)[/tex]

[tex]\text{Double Angle:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{2(A+B)}{2(2)}\bigg)\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+2\sin \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)[/tex]

[tex]\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg][/tex]

[tex]\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg][/tex]

[tex]\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)[/tex]

[tex]\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{\pi -C}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)[/tex]

LHS = Middle [tex]\checkmark[/tex]

Proof Middle → RHS:

[tex]\text{Middle:}\qquad 4\cos \bigg(\dfrac{\pi -A}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -C}{4}\bigg)\\\\\\\text{Given:}\qquad \qquad 4\cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)\cdot \cos \bigg(\dfrac{A+B}{4}\bigg)\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{B+C}{4}\bigg)\cdot \cos \bigg(\dfrac{C+A}{4}\bigg)[/tex]

Middle = RHS [tex]\checkmark[/tex]

Answer:

Step-by-step explanation:10

Answer:

12

Step-by-step explanation:

4x3=12

Answer:13

Step-by-step explanation:

13^2

Answer:

see below

Step-by-step explanation:

 g(x) = 1/2 | x+2| -3

y = Cf(x)   0 < C < 1 compresses it in the y-direction

y = f(x + C)      C > 0 moves it left

y = f(x) + C C < 0 moves it down

1/2 compression in the y direction moved to the left 2 units and down 3 units

Given: g(x) = 1/2|x + 2| - 3

0 < c < 1 means at the y-axis and makes it smaller

y = cf(x)

1/2 compression

c > 0 means that you move it to the left

y = f(x + c)

2 units to the left

c < 0 means that you move it down

y = f(x) + c

3 units down

Best of Luck!

Answer:

m∠XAZ = 40°

Step-by-step explanation:

Since ∠YAU, ∠YAZ and ∠XAZ are the angles based at a point on a straight line, and the sum of the all angles on a line is 180°.

Therefore, m∠YAU + m∠YAZ + m∠XAZ = 180°

50° + 90° + m∠XAZ = 180°

m∠XAZ = 180° - 140°

             = 40°

Therefore, measure of angle XAZ is 40°.

Answer: The answer is 0.11 in decimal form.

Step-by-step explanation:

Answer: False

Step-by-step explanation: First, you are going to multiply the numbers in the first set of parenthesis, getting 72. Now, have 72 take the place of 12x6 in the equation. Your new equation looks like this: 0.3 x 72 = 0.4(12-3).

Going off of step one, you will be doing the same thing with the second set, but you will be subtracting instead. You will get 9, and plug it place of the parenthesis. Your equation now looks like o.3 x 72 = 0.4 x 9.

Next, you will multiply the set of numbers on the left side of the equal sign, getting 21.6. You now have an equation of 21.6 = 0.4 x 9.

Going off of step 3, you are going to do the exact same thing we did in step 3. Multiply 0.4 and 9, and you will get 3.6. The equation now looks like 21.6 = 3.6.

Now, the inequality is false because the left-hand side and right-hand side are two completely different numbers. A bigger number cannot be equal to a smaller number.

Hope this helps!

Answer:

A, D, E

Step-by-step explanation:

like with division if you multiply the same thing on both sides it is still equal

The  ratios are equivalent to 8 : 12 are A 4 : 6 , D 24 : 36 and E 56 : 84.

Those expressions who might look different but their simplified forms are same expressions are called equivalent expressions.

Suppose that we've got two quantities with measurements as 'a' and 'b'

Then, their ratio(ratio of a to b) a:b

We usually cancel out the common factors from both the numerator and the denominator of the fraction we obtained. Numerator is the upper quantity in the fraction and denominator is the lower quantity in the fraction).

Suppose that we've got a = 8, and b= 12,

[tex]a:b = 8 : 12\\\\ = \dfrac{8}{12} = \dfrac{2 \times 4}{2 \times 6} \\= \dfrac{2}{3} \\\\a : b = 2 : 3[/tex]

So, the  ratios are equivalent to 8 : 12

A 4 : 6

D 24 : 36

E 56 : 84

Remember that the ratio should always be taken of quantities with same unit of measurement. Also, ratio is a unitless(no units) quantity.

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Answer:

[tex]x<-8[/tex]

Step-by-step explanation:

So we have the inequality:

[tex]3x-1>4x+7[/tex]

Add 1 to both sides. The left side cancels:

[tex]3x>4x+8[/tex]

Subtract 4x from both sides. The right side cancels:

[tex]-x>8[/tex]

Divide both sides by -1. Since we're dividing by a negative, we flip the sign:

[tex]x<-8[/tex]

And we're done!

Our solution is all numbers less than -8.

Answer:

x = -8

Step-by-step explanation:

I will be the brainliest, its hard to explain on a document, i could make you avideo how to do these problems, im a math teacher.

We need to multiply the 80mph by the 1.6 to make it KPH

When done correctly we get the answer of 128KPH

The limit is 128kph

HOPE THIS HELPED. DONT FORGET TO MARK THIS AS THE BRAINLIEST ANSWER AND GIVE IT A 5 STAR SO OTHER PEOPLE CAN BE HELPED BY THIS ANSWER!! HAVE A NICE DAY!

Answer:

56

Step-by-step explanation:

Here, n = 8, r = 3

[tex] \huge \orange{ \because \: ^nC_r = \frac{n!}{r!(n - r)!} } \\ \\ \huge \therefore \: ^8C_3 = \frac{8!}{3!(8 - 3)!} \\ \\ \huge \therefore \: ^8C_3 = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} \\ \\ \huge \red{ \boxed{\therefore \:^8C_3 = \purple{56}}}[/tex]

Answer:

40

Step-by-step explanation:

because if you do 400 divided by 10, it gives you 40

Answer:

His average speed is 4.75 m/s

Step-by-step explanation:

Here, we are interested in calculating the average speed of the boy.

At time t = 4 seconds, total distance traveled = 6 meters

The average speed here is distance/time = 6/4 = 1.5 m/s

Now at time 30 7 seconds , he has rolled 30 meters

What this actually means is that for the next 3 seconds after the initial 4 seconds, he rolled a total distance of 30-6 = 24 meters. Hence, he rolled 24 meters in the next three seconds

So mathematically, his average speed for the second part of the rolling is 24/3 = 8 meters per second

Conclusively therefore , his average speed will be (1.5 + 8)/2 = 9.5/2 = 4.75 meters per second

answer:

n=3

Step-by-step explanation:

6n-10+3n=17

6n+3n=27

9n=27

n=3

Answer:

B) (-2,3)

Step-by-step explanation:

Graph all your points. No count the distance from point A to B. divide in half .

5 1.80, the crackers cost $1.80 per box: 20 3 1.80 5 36. So, 20 boxes cost $36.00.

Answer:

$36

Step-by-step explanation:

To find how much one box costs you divide 9÷5 = 1.8

Then you take 1.8 × 20 = 36

If you want me to go over it in depth, let me know! Have a great day!

Answer:

1 1/3

Step-by-step explanation:

Firstly, 1 2/3 is converted into an improper fraction being 5/3. Next, 1/5 of 5/3 is one portion of 5 equal portions. You can split 5/3 into five equal portions of 1/3. Therefore, 1/3 cup of honey was used to make the honey glaze. Lastly, that is subtracted from the 1 2/3 to end up with 1 1/3.

The required amount of honey left is given as 1 1/3.

Given that,
Ben wants to try a new honey-glazed salmon recipe. He starts with 1 2/3 cups of honey. He uses 1/5 of it to make the glaze.

In mathematics, it deals with numbers of operations according to the statements. There are four major arithmetic operators, addition, subtraction, multiplication and division,

The process in mathematics to operate and interpret the function to make the function or expression simple or more understandable is called simplifying and the process is called simplification.

here,
Amount of honey = 1 2/3 = 5 / 3
Amount of honey used is 1/5 times = 1/5 [5 / 3] = 1 / 3
Amount of honey left = 5/3 - 1/3 = 4/3 or 1 1/3

Thus, the required amount of honey left is given as 1 1/3.

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[tex] \huge \purple {\tt {x + \frac{7}{6} = 2x - \frac{7}{6}}} [/tex]

[tex] \large {\tt {The \: answer \: in \: reduced}} [/tex]

[tex] \large {\tt { improper \: fraction \: form.}} [/tex]

[tex] \huge \pink {\tt {x + \frac{7}{6} = 2x - \frac{7}{6}}} [/tex]

[tex] \huge \blue {\tt {or, \: x - 2x = - \frac{7}{6} - \frac{7}{6}}} [/tex]

[tex] \huge \red {\tt {or, \: - x = \frac{ - 14}{6} = \frac{ - 7}{3}}} [/tex]

[tex] \huge \green {\tt {or, \: x = \frac{7}{3}}} [/tex]

[tex] \huge \orange {\tt {x = \frac{7}{3}}} [/tex]