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2 years ago

Select the true statement about triangle ABC.

Mathematics

1 answer:

almond37 [142]2 years ago

8 0

Answer:

Step-by-step explanation:

cosA = $\frac{adjacent}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{12}{13}$

cosC = $\frac{BC}{AC}$ = $\frac{5}{13}$

sinC = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{12}{13}$

tanC = $\frac{opposite}{adjacent}$ = $\frac{AB}{BC}$ = $\frac{12}{5}$

Then

cosA = sinC → B

You might be interested in

Word phrase for 6a-2b

kirill115 [55]

6a - 2b

six times a number decreased by 2 times another number

3 0

3 years ago

Rewrite in simplest radical form, thx :)

Aleks [24]

Answer:

The answer is

$\sqrt[3]{ {x}^{2} }$

Step-by-step explanation:

$\frac{ {x}^{ \frac{5}{6} } }{ {x}^{ \frac{1}{6} } }$

Since they have the same base and are dividing we can use the rules of indices

That's

subtract the exponents

So we have

${x}^{ \frac{5}{6} - \frac{1}{6} }$

${x}^{ \frac{4}{6} }$

$= {x}^{ \frac{2}{3} }$

Rewriting it in radical form

We have the final answer as

$\sqrt[3]{ {x}^{2} }$

Hope this helps you

6 0

2 years ago

Help me with this question I’m confused

MAXImum [283]

Answer:

3/4 is the slope

Step-by-step explanation:

I dont really know how to explain

7 0

2 years ago

Calculate AD and hence BD.

Vilka [71]

Answer:

AD = 15 cm , BD = 3 cm

Step-by-step explanation:

Δ ABC is similar to Δ ADE ( by the AA postulate )

then the ratios of corresponding sides are in proportion, that is

$\frac{AD}{AB}$ = $\frac{DE}{BC}$ ( substitute values )

$\frac{AD}{12}$ = $\frac{10}{8}$ ( cross- multiply )

8 × AD = 12 × 10 = 120 ( divide both sides by 8 )

AD = 15 cm

AB + BD = AD , that is

12 + BD = 15 ( subtract 12 from both sides )

BD = 3 cm

7 0

1 year ago

Will give brainliest to best answer.

Tomtit [17]

Problem One
Remark
If she doesn't mind having I kg left over, the minimum number would be 3 five kg boxes. If on the other hand, she must have exactly 14 kg then the minimum number is 6.

She needs 2 five kg boxes and 4 one kg boxes. <<<< Answer

Problem Two
There is a method of solving this that is called dimensional analysis. It is what should be used here. I'll do it at the end of the problem. In the meantime, you have to do it a slightly longer way.

1 portion = 100 grams.
x portions = 1kg which is 1000 grams.
x portions = 1000 grams.

Set up a proportion to find the number of servings in 1 kg
1 portion/x = 100 grams/1000 grams Cross multiply
1 * 1000 = 100 * x Divide by 100
1000/100 = x
x = 10 servings in 1 kg.

So each kg produces 10 portions
1 kg / 10 portions = 20 kg / x portions Cross multiply
x * 1 = 10 * 20
x = 200 portions <<<<< Answer

Dimensional Analysis
[1batch]*[1 portion/100g][1000g/kg][20kg/batch] the units cancel
1000 * 20 / 100 only the portions are left over.
200 portions is the answer.

Problem Three
1 kg = 1000 grams
x kg = 5000 grams Cross multiply

1*5000 = 1000 x
x = 5 kg

1 parcel weighs 5 kg
x = 15 kg

15 kg = 5 x
x = 15/5
x = 3

So he can carry 3 parcels per trip.

Since there are 5 such parcels, he will have to make 2 trips. The second one will not be a full load.
First Trip = 3 parcels
Second Trip = 2 parcels. <<<<Answer

5 0

2 years ago