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3 years ago

Help pretty pretty please :(

Mathematics

2 answers:

Evgen [1.6K]3 years ago

8 0

Answer:

27/20

Step-by-step explanation:

The absolute value of any expression is always positive or zero

3/5 + 3/4

Add the fractions

27/20

Solution

27/20

Alternate form

1 7/20, 1.35

posledela3 years ago

6 0

Answer:

1.35

Step-by-step explanation:

Simplify: 3/5 + 3/4
Find the LCM for the denominators: 20
Alter the numerators to fit the denominators: 3 × 4 = 12, 3 × 5 = 15
Re-write the fractions and add them: $\frac{12}{20} + \frac{15}{20} = \frac{27}{20}$
27/20 = 1.35

I hope this helps!

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Answer: First option.

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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4 years ago

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givi [52]

Answer:

Step-by-step explanation:

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3 years ago

A person is standing on a cliff that is 200ft above a body of water. The person looks down at an angle of depression of 42o at a

velikii [3]

Answer:

524.5 feet

Step-by-step explanation:

Given: Height of cliff= 200 ft

Angle of depression at sail boat is 42°

Angle of depression at yacht is 15°

Lets assume distance sailboat from the bottom of cliff is " $d_1$ "

And assume distance yacht from the bottom of cliff is " $d_2$ "

Now using tangent rule to solve it.

we know, $tan\theta = \frac{opposite}{adjacent}$

Distance sailboat from the bottom of cliff; $tan 42= \frac{200}{d_1}$

Using trignometry table to know the value of tan 42°

⇒ $0.90= \frac{200}{d_1}$

cross multiplying both side.

⇒ $d_1= \frac{200}{0.90} = 222.22 \approx 222$

∴ Distance sailboat from the bottom of cliff ( $d_1)$ = 222 feet

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Using trignometry table to know the value of tan 15°

⇒ $0.2679= \frac{200}{d_2}$

cross multiplying both side.

⇒ $d_2= \frac{200}{0.2679} = 746.54 \approx 746.5$

∴Distance yacht from the bottom of cliff $(d_2)$ = 746.5 feet.

Next, finding the distance between sailboat and yacht.

Distance between sailboat and yacht = $d_2-d_1$

⇒ Distance between sailboat and yacht= $746.5-222= 524.5\ ft$

∴ Distance between sailboat and yacht is 524.5 feet.

3 0

3 years ago