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3 years ago

randy just had a lap pool put in his backyard. The rectangle pool measures 5 feet by 18 feet and is 4.5 feet deep

Mathematics

1 answer:

photoshop1234 [79]3 years ago

4 0

Do 5 times 8 then divide 4.5

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Ms.tang bought 4 sets of doughnuts. she created a package of treats using 2/3 of a set. how many package did she created using a

SSSSS [86.1K]

Answer:

Step-by-step explanation:

2/3 times 4 sets is 8/3, which simplifies to 2 1/3. but you cant have 1/3 of a package so you round to 2

8 0

3 years ago

Which equation has a solution of (-2,1)?

Vlada [557]

C because you have to substitute so 1=-2(-2)-3

7 0

3 years ago

In ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. Find the measure of ∠J to the nearest degree.

nydimaria [60]

We have been given that in ΔJKL, the measure of ∠L=90°, KL = 22 feet, and JK = 54 feet. We are asked to find the measure of angle J to nearest degree.

First of all, we will draw a triangle as shown in the attachment.

We can see from our attachment that side KL is opposite side to angle J and side JK is hypotenuse of right triangle.

We know that sine relates opposite side of right triangle to hypotenuse.

$\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}$

$\text{sin}(\angle J)=\frac{22}{54}$

Using inverse sine or arcsin, we will get:

$\angle J=\text{sin}^{-1}(\frac{22}{54})$

$\angle J=24.042075905756^{\circ}$

Upon rounding to nearest degree, we will get:

$\angle J\approx 24^{\circ}$

Therefore, the measure of angle J is approximately 24 degrees.

8 0

4 years ago

What is 4c ?? Answer ASAP pls!!

slava [35]

Answer:

Step-by-step explanation:

Multiply 4 to each number

5 0

3 years ago

Suppose that S is the set of successful students in a classroom, and that F stands for the set of freshmen students in that clas

lora16 [44]

Answer:

b) 24

Step-by-step explanation:

We solve building the Venn's diagram of these sets.

We have that n(S) is the number of succesful students in a classroom.

n(F) is the number of freshmen student in that classroom.

We have that:

$n(S) = n(s) + n(S \cap F)$

In which n(s) are those who are succeful but not freshmen and $n(S \cap F)$ are those who are succesful and freshmen.

By the same logic, we also have that:

$n(F) = n(f) + n(S \cap F)$

The union is:

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

In which

$n(S \cup F) = 58$

$n(s) = n(S) - n(S \cap F) = 54 - n(S \cap F)$

$n(f) = n(F) - n(S \cap F) = 28 - n(S \cap F)$

$n(S \cup F) = n(s) + n(f) + n(S \cap F)$

$58 = 54 - n(S \cap F) + 28 - n(S \cap F) + n(S \cap F)$

$n(S \cap F) = 24$

So the correct answer is:

b) 24

3 0

4 years ago