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2 years ago

Which of the following statements is not true?

Mathematics

2 answers:

Thepotemich [5.8K]2 years ago

7 0

The above question is asking us to check if the exponents are written in the correct way or not.Let us look at the options one by one.

1) The exponent denotes the number of time the letter or number is multiplied.We have six x which are multiplied so the exponent for x which is 6 is right .The exponent is written correct .

2)We have seven A which are multiplied and five B multiplied making the exponent of A as 7 and B as 5 .The exponents are written correct.

3)There are six b and six c .The exponent rule has been followed here that is we can keep bc inside parenthesis and make 6 as the exponent.

4)Here exponent for x,y,z all are 5. we can not remove the exponent of x and y .It is not following the exponent rule .

Option 4 is not true.

Amiraneli [1.4K]2 years ago

4 0

The above question is asking us to check if the exponents are written in the correct way or not.Let us look at the options one by one.

1) The exponent denotes the number of time the letter or number is multiplied.We have six x which are multiplied so the exponent for x which is 6 is right .The exponent is written correct .

2)We have seven A which are multiplied and five B multiplied making the exponent of A as 7 and B as 5 .The exponents are written correct.

3)There are six b and six c .The exponent rule has been followed here that is we can keep bc inside parenthesis and make 6 as the exponent.

4)Here exponent for x,y,z all are 5. we can not remove the exponent of x and y .It is not following the exponent rule .

Option 4 is not true.

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Find sec A and cot B exactly if a = 8 and b = 7.

Shtirlitz [24]

Answer:

sec A= 1.01 and cot B =8.25

Step-by-step explanation:

Given :

sec A and cotB if a =8 and b=7

Now,

= $sec A=\frac{1}{cos A} \\\\\frac{1}{cos 8} \\\\\frac{1}{0.99} \\1.01$

and

$cot B=\frac{cosB}{sin B} \\cotB=\frac{cos 7}{sin7} \\cot B =\frac{0.99}{0.12} \\cot B =8.25$

Therefore, answer will be sec A= 1.01 and cot B =8.25

8 0

2 years ago

Lowest common multiple of 12,18,56

Ira Lisetskai [31]

Step-by-step explanation:

12 = 2 × 2 × 3

18 = 2 × 3 × 3

56 = 2 ×2 × 2 × 7

Now

Common factor = 2

Remaining factor = 2 × 3 × 2 × 3 × 7

LCM = RF × CF

= 504

hence the lCM of 12 , 18 and 56 is 504...

$...$

8 0

1 year ago

Read 2 more answers

If the area of AABC is D, give the expressions that complete the equation for the measure of ZB?

Ira Lisetskai [31]

Given:

The figure of triangle ABC.

The area of the triangle ABC is D.

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$

To find:

The value of m and n in the given expression.

Solution:

Let h be the height of the triangle ABC.

Area of a triangle is:

$Area=\dfrac{1}{2}\times base\times h$

Where, b is the base and h is the height of the triangle.

$Area=\dfrac{1}{2}\times a\times h$

The area of the triangle ABC is D.

$D=\dfrac{1}{2}\times a\times h$

$2D=ah$

$\dfrac{2D}{a}=h$ ...(i)

In a right angle triangle,

$\sin \theta =\dfrac{Perpendicular}{Hypotenuse}$

$\sin B =\dfrac{h}{c}$

$\sin B =\dfrac{1}{c}\times \dfrac{2D}{a}$ [Using (i)]

$\sin B =\dfrac{2D}{ac}$

$m\angle B =\sin ^{-1}\dfrac{2D}{ac}$ ...(ii)

We have,

$m\angle B=\sin ^{-1}(\dfrac{m}{n})$ ...(iii)

On comparing (ii) and (iii), we get

$m=2D$

$n=ac$

Therefore, the required values are $m=2D, n=ac$ .

6 0

2 years ago

14) Solve for x: 4 + 2(x + 3x) = 4 - x(2 - 3) + 2x A) 0 B) 1 2 D) 3

tester [92]

answer;

x=0

Step-by-step explanation:

4 + 2 (x + 3x) = 4 - x(2-3) + 2x

let remove d brackets

4 + 2x + 6x = 4 - 2x + 3x + 2x

llet collect like terms

2x + 6x - 3x -2x + 2x = 4-4

5x = 0

x = 0

3 0

2 years ago

Asking for help in the following question #1 and #2 are based of algebra 2. ≅ ⇒⇒⇒⇒⇒⇒

Orlov [11]

Answer:

Step-by-step explanation:

For the first one, revenue - cost = profit. We have equations for revenue and profit, so filling in:

$-.3x^2+150x-cost=-.5x^2+250x-300$

Let's add cost to both sides to make it positive and bring everything on the right over to the left and combine like terms:

$.2x^2-100x+300=cost$

For the second one:

P(x)*T(x) = $(x^2-3x-7)(3x)$

Distribute the 3x into everything inside the parenthesis to get

$3x^3-9x^2-21x$

For the second part of that problem, C(x)*P(x) = $(x-4)(x^2-3x-7)$

Distribute the x into everything first to get:

$x^3-3x^2-7x$

The distribute the -4 into everything to get:

$-4x^2+12x+28$

Combine the like terms and put everything together to get

$x^3-7x^2+5x+28$

7 0

2 years ago