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Ierofanga [76]
3 years ago
13

I’am between 750 and 850. I have 3 ones and no tens.

Mathematics
1 answer:
il63 [147K]3 years ago
7 0
Maybe something like 811.1?
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Take 9.5 gallons 6 1/3 poured out how much water is left in the tank
elena55 [62]
We can first convert 9.5 into fraction :

9.5= 9 1/2

Then we can use deduction to find out the answer:

9 1/2 - 6 1/3

=19/2 - 19/3

We change it to the same denominator:

=19×3/6 - 19×2/6

= 57/6 - 38/6

= 19/6

Therefore there are 19/6 or 3 1/6 gallons left in the tank.

Hope it helps!
4 0
3 years ago
Please help and thank you
mamaluj [8]

Answer:

C

Step-by-step explanation:

a is defined first as 5. So a = 5.

b is defined after the circle b = -2

a^2 = 5^2 = 25

B^4 =(- 2)^4 = 16

Therefore sqrt(a^2 * b^4) = sqrt(25 * 16) = sqrt(400) = 20

The answer is C

5 0
2 years ago
Read 2 more answers
There are 7 days in 1 week. How many days are there<br> in 4 weeks?
otez555 [7]

Answer:

28 days

Step-by-step explanation:

You multiply 7 by 4 for find the number of days in 4 weeks, since 1 week is 7 days: 7 × 4 = 28.

Hope this helps!

8 0
2 years ago
Read 2 more answers
ΔABC is dilated by a scale factor of 0.25 with the origin as the center of dilation, resulting in the image ΔA′B′C′. If A = (-1,
Leni [432]

Answer:

5

Step-by-step explanation:

To find B and C prime, you must multiply them by .25, or 1/4.


B' =  

(-2 x .25),(1 x .25)


I did mine in fraction form, because it will prove to be more useful in future mathematics.


B' = (1/2  ,  1/4)


Repeat the process with C.


C' =

(14 x .25),(17 x .25)


C' =

(7/2  ,  17/4)




You only need to focus on B and C because you are finding the length of B'C'.


The formula for distance is the square root of x to the sub of 2 minus x to the sub of 1 squared minus y to the sub of 2 minus y to the sub of 1 square.


x2 -  x1 =  7/2  -  1/2  =  6/2  =  3 squared  =  9


y2 - y1 = 17/4 - 1/4 = 16/4  =  4 squared = 16


16 + 9  = 25


Square root of 25 is 5.


Therefore, the distance is 5.


8 0
3 years ago
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
2 years ago
Read 2 more answers
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