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3 years ago

Solve x - 6 = 3 Question 3 options: x = 3 x = 2 x = 9 x = -3

Mathematics

2 answers:

Agata [3.3K]3 years ago

6 0

Answer:

x = 9

Step-by-step explanation:

Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

x-6-(3)=0

Step by step solution :

Step 1 :

Solving a Single Variable Equation :

1.1 Solve : x-9 = 0

Add 9 to both sides of the equation :

x=9

brainly pleasse :)

Arte-miy333 [17]3 years ago

5 0

Answer:

x = 9

Step-by-step explanation:

x - 6 = 3

Add 6 to both sides

x - 6 + 6 = 3 + 6

x = 9

I hope this was helpful, please mark as brainliest

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Use rounding or compatible numbers to estimate each product or quotient.

zmey [24]

Part 1

Using rounding or compatible numbers, we estimate the product

$5\times2 \frac{3}{5}$
to
$5\times3$

Therefore, the estimate of the product $5\times2 \frac{3}{5}$ , using rounding and compatible numbers is about 15.

Part 2

Using rounding or compatible numbers, we estimate the quotient

$25\frac{7}{8}\div2 \frac{8}{9}$
to
$27\div3$

Therefore, the estimate of the quotient $25\frac{7}{8}\div2 \frac{8}{9}$ , using rounding and compatible numbers is about 9.

Part 3

Given that Mr. Garcia's class ate $3\frac{2}{3}$ pizzas, Mrs Rizzoli's class ate $4\frac{1}{8}$ pizzas, and Mrs. Tindal's class ate $3\frac{1}{4}$ pizzas.

Using benchmarks, we estimate the total number of pizzas eaten by all three classes as follows: $3\frac{3}{4}$ , $4$ and $3\frac{1}{4}$

Thus, the total number of pizzas eaten by all three classes is given by

$3\frac{3}{4}+4+3\frac{1}{4}=11$

Therefore, the total number of pizzas eaten by all three classes is about 11.

3 0

4 years ago

Brainliest if correct! Please explain how this is done!

Korvikt [17]

The value of csc C in fraction and in decimal form is as follows:

csc C = 5 / 3

csc C = 1.66666666667

Using trigonometric ratio,

secant A = 1 / cos A = hypotenuse / adjacent

Therefore,

sec A = 5/3

sec A = 1 / cos A

cos A = adjacent / hypotenuse

cos A = 3 / 5

A = cos⁻¹ 0.6

A = 53.1301023542

Therefore,

∠A = 53°

∠B = 90°

∠C = 180 - 143 = 37°

cosec C = 1 / sin C = hypotenuse / opposite

cosec C = 5 / 3

learn more about trigonometry here: brainly.com/question/25942962?referrer=searchResults

5 0

2 years ago

Help me please I need

vagabundo [1.1K]

Answer:

I think the answer is A=49

8 0

3 years ago

If y=2×-3, determine the value of y when ×=-15

Harrizon [31]

Answer:

-33

Step-by-step explanation:

If y=2× - 3 ,when ×=-15

Y= 2(-15) - 3

Y = -30-3

Y = -33

6 0

3 years ago

The accompanying data came from a study of collusion in bidding within the construction industry. No. Bidders No. Contracts 2 6

viktelen [127]

Answer:

The proportion of contracts involved at most five bidders is 0.667.

The proportion of contracts involved at least five bidders is 0.51.

The proportion of contracts involved between five and 10 inclusive bidders is 0.5.

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

Step-by-step explanation:

No. bidders No. contracts Relative frequency of contracts

2 6 6/102=0.0588

3 20 20/102=0.1961

4 24 24/102=0.2353

5 18 18/102=0.1765

6 13 13/102=0.1275

7 7 7/102=0.0686

8 5 5/102=0.049

9 6 6/102=0.0588

10 2 2/102=0.0196

11 1 1/102=0.0098

Total 102

We have to find proportion of contracts involved at most five bidders.

Proportion of at most 5= Relative frequency 2+ Relative frequency 3+ Relative frequency 4+ Relative frequency 5

Proportion of at most 5=0.0588+ 0.1961+0.2353+0.1765

Proportion of at most 5=0.6667

The proportion of contracts involved at most five bidders is 0.667.

proportion of at least five bidders= proportion≥5= 1- proportion less than 5

Proportion less than 5=0.0588+ 0.1961+0.2353=0.4902

proportion of at least five bidders=1-0.4902=0.5098

The proportion of contracts involved at least five bidders is 0.51

We have to find proportion of contracts involved between five and 10 inclusive bidders.

Proportion of contracts between five and 10 inclusive= Relative frequency 5+ Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9+ Relative frequency 10

Proportion of between five and 10 inclusive=0.1765 +0.1275 +0.0686 +0.049 +0.0588 +0.0196

Proportion of between five and 10 inclusive=0.5

The proportion of contracts involved between five and 10 inclusive bidders is 0.5

We have to find proportion of contracts involved strictly between five and 10 bidders.

Proportion of contracts strictly between five and 10= Relative frequency 6+ Relative frequency 7+ Relative frequency 8+ Relative frequency 9

Proportion of strictly between five and 10=0.1275 +0.0686 +0.049 +0.0588

Proportion of strictly between five and 10=0.3039

The proportion of contracts involved strictly between five and 10 bidders is 0.304.

8 0

3 years ago