Answer:
Question 1 (1 point)
1. An object has a mass of 25g and a volume of 5ml
a
10
b
8
c
5
d
7
Question 2 (1 point)
2. An object has a mass of 30g and a volume of15ml. What is the density?
a
10
b
2
c
3
d
5
Question 3 (1 point)
3. An object has a mass of 200g and a volume of 100ml. What is the density?
a
2
b
4
c
6
d
8
Question 4 (1 point)
4. An object has a mass of 17g and a volume of 2ml. What is the density?
a
2.4
b
2
c
8
d
8.5
Question 5 (1 point)
5. An object has a mass of 20g and a volume of 5ml. What is the density?
a
2
b
4
c
10
d
2.5
Question 6 (1 point)
6. An object has a mass of 75g and a volume of 75ml. What is the density?
a
6
b
1
c
7
d
8
Question 7 (1 point)
7. An object has a volume of 100ml and a mass of 2oo ml. What is the density?
a
0.5
b
1
c
4
d
2.5
Question 8 (1 point)
8. An object has a volume of 5ml and a mass of 650 g. What is the density?
a
110
b
120
c
130
d
200
Question 9 (1 point)
9. An object has a mass of 15g and a volume of 5ml. What is the density?
a
2.5
b
7
c
3
d
2.56
Question 10 (1 point)
10. An object has a mass of 13g and a volume of 2ml. What is the density?
a
6.5
b
3.7
c
4
d
Step-by-step explanation:
Answer:
x ≈ 16.5
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry</u>
- [Right Triangles Only] SOHCAHTOA
- [Right Triangles Only] cos∅ = adjacent over hypotenuse
Step-by-step explanation:
<u>Step 1: Define</u>
We are given a right triangle. We can use trig to find the missing side length.
<u>Step 2: Identify Variables</u>
<em>POV from the angle measure</em>
Angle = 38°
Adjacent Leg = <em>x</em>
Hypotenuse = 21
<u>Step 3: Solve for </u><em><u>x</u></em>
- Substitute [cosine]: cos38° = x/21
- Isolate <em>x</em>: 21cos38° = x
- Rewrite: x = 21cos38°
- Evaluate: x = 16.5482
- Round: x ≈ 16.5
Answer:
144
Step-by-step explanation:
Plug in a and b
In this case a=4, b=3
(ab)^2
( (4)(3) )^2
(12)^2
12x12
144
If this helps please mark as brainliest
The zeroes of the function are 
.
Solution:
Given function:
<u>To find the zeros of the function:</u>
⇒ f(x) = 0
Take x² as common in first bracket and take 12 as common in 2nd bracket.
Now, take common term (x + 3) outside.
<em>Using zero factor principle, If ab = 0 then a = 0 or b = 0.</em>
x = –3
x² = –12
x² = 2² × 3 × –1
Taking square root on both sides, we get
we know that 
.
Hence the zeroes of the function are .
Ok so basically what u need to do is
