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3 years ago

9

What is visible by 18 and 23

1 answer:

Tasya [4]3 years ago

6 0

The only number I got was 1. I don’t think you divide 23 unless you want the remainder

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Mittens, a baby kitten, gained 120 grams of weight in 1/3 of a month. If Mittens kept growing at this rate, how much weight woul

dlinn [17]

Answer:

Step-by-step explanation:

Given

A baby kitten, gained 120 grams of weight in 1/3 of a month.

he gain in 1 month

Solution

1/3 month =120gram weight

1 month =120×3

=360 gram is the answer

6 0

2 years ago

Can you help me plz <br><br> take your time tho

AnnZ [28]

Answer:

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Step-by-step explanation:

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2 years ago

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7/9 in simplest form

ozzi

Find the GCD (or HCF) of numerator and denominator
GCD of 7 and 9 is 1Divide both the numerator and denominator by the GCD
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3 years ago

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HELP MEEEEEEEEE plz

Diano4ka-milaya [45]

Answer:

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Step-by-step explanation:

Red : Blue :

3 8

We have 40 blue marbles

Multiply 8 by what to get 40

40/8 =5

So we need to multiply by 5

Red : Blue :

3*5 8*5

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3 years ago

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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

3 years ago

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