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Alborosie
3 years ago
7

Solve for b and explain step by step please: 12b=8b+16

Mathematics
2 answers:
Kobotan [32]3 years ago
8 0
I hope this helps if not and have any questions just ask me

Gemiola [76]3 years ago
5 0
12b-8b=16
4b=16/4
b=4

therefore b is 4
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If 3(x+2)=5(x-8) what is the value of x+2?<br><br>A. 23<br>B.25<br>C.40<br>D.46
lorasvet [3.4K]

3*x

3*2

3x+6

5*x

5*-8

5x-40

Story short the answer is 25

3 0
3 years ago
Read 2 more answers
Find the area of this parallelogram?
Jet001 [13]

So, the area of a parallelogram is the base times the height or

a = b \times h

a being area

b being base

h being height

the height is 8 cm and the base is 15 cm

(remember: the heght should be perpendicular to the base)

so we plug in 8 for h and 15 for b to get

a = 15 \times 8

and 15 × 8 = (10 × 8) + (5 × 8) = 80 + 40 =120

So the answer is 120

3 0
3 years ago
If Tom runs 3.5 miles in 29.4 minutes, how many hours would it take him to run 26 miles? Show your work.
Thepotemich [5.8K]

Answer:

Step-by-step explanation:

3.5 miles/(29.4 minutes)

26 miles × (29.4 minutes)/(3.5 miles) = 218.4 minutes

218.4 minutes × (1 hour)/(60 minutes) = 3.64 hours

4 0
3 years ago
Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.
galina1969 [7]

Answer:

All potential roots are 3,3 and -\frac{5}{3}.

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

f(x)=3x^3-13x^2-3x+45

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45

q=\pm 1,\pm 3

All possible roots

\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}

Now we check each rational root and see which are possible roots for given function.

f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0

f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32

f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144

f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}

Similarly, we will check for all value of p/q and we get

f(-5/3)=0

Thus, All potential roots are 3,3 and -\frac{5}{3}.


5 0
3 years ago
Read 2 more answers
Some body can help me with a geometric mean maze
Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.

2. South-East point from the Start: By the 2nd theorem,

x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.

3. West point from the previous: By the 2nd theorem,

x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.

4. West point from the previous: By the 1st theorem,

9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.

5. West point from the previous: By the 2nd theorem,

10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.

6. North point from the previous: By the 1st theorem,

x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.

7. East point from the previous: By the 2nd theorem,

x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.

8. North point from the previous: By the 1st theorem,

x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.

8. West point from the previous: By the 2nd theorem,

x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.

9. North point from the previous: By the 1st theorem,

12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.

101. East point from the previous: By the 1st theorem,

6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.

11. East point from the previous: By the 2nd theorem,

20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.

12. South-east point from the previous: By the 2nd theorem,

18^2=x\cdot 21.6\Rightarrow x=15.

13. North point=The end.

6 0
3 years ago
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