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Anika [276]
3 years ago
8

Solving a One-Variable Linear Equation

Mathematics
1 answer:
vitfil [10]3 years ago
5 0

Answer:

\Longrightarrow:\boxed{\sf{h=14}}

Step-by-step explanation:

Isolate the term of h from one side of the equation.

Use the distributive property.

<u>Distributive property:</u>

→ A(B+C)=AB+AC

<h3>7h-5(3h-8)=-72</h3>

<u>First, multiply expand.</u>

⇒ -5(3h-8)

⇒ -5*3h=-15h

⇒ -5*8=40

<u>Rewrite the problem.</u>

→ -15h+40

⇒ 7h-15h+40=-72

⇒ 7h-15h=-8h

→ -8h+40=-72

<u>Subtract by 40 from both sides.</u>

⇒ -8h+40-40=-72-40

<u>Solve.</u>

⇒ -72-40=-112

⇒ -8h=-112

<u>Then, you divide by -8 from both sides.</u>

⇒ -8h/-8=-112/-8

<u>Solve.</u>

<u>Divide the numbers from left to right.</u>

→ -112/-8=14

\Longrightarrow: \boxed{\sf{h=14}}

  • <u>Therefore, the correct answer is h=14.</u>

I hope this helps! Let me know if you have any questions.

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Heeeeeeeeeeeeeeeeeeepl
DIA [1.3K]

Answer:

Hi there!

Your answer is;

a)

i) 400% of 240

240 is 100%

× 4

960= 400%

ii) 40% of 240

100% = 240

/100

1% = 2.4

× 40

40% = 96

iii) 4% of 240

100% is 240

/100

1% = 2.4

× 4

4% = 9.6

iv) .04% of 240

100% = 240

/100

1% = 2.4

/100

.01% = .024

× 4

.04% = .096

b) the patterns is that all these numbers equal sometime 96. Each of these have a different decimal place, but have the same actual numbers.

c) 4000% = 240

take the pattern:

400% is 960

Scale it up to 4000 by 10

400% is 960

× 10

4000% is 9600

Hope this helps!

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3 years ago
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10 students ride the bus.
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3 years ago
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Wewaii [24]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At
kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

(a) So, the probability of B,

P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

\implies P(A\cap B\cap C)=\frac{1}{8}

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3 years ago
Alex has a grid in the shape of a rectangle. He has right triangles. Each right triangle has an area of 25 square millimeters an
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First I would convert 20 and 12 to millimeters which would be 20= 200mm 12= 120mm. You would then find the area if the grid which would be 200•120=
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3 years ago
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