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2 years ago

In the expression 14x + 3x – 3y + 14, which terms are “like terms”?

Mathematics

1 answer:

vfiekz [6]2 years ago

7 0

Answer:

14x and 3 x

Step-by-step explanation:

14x and 3 x are like terms.

This is because they both have 'x'. (They can even be simplified)

Whereas unlike terms are terms which do not contain the same variable or numerical value. (They cannot be simplified with other terms than its like term)

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100 kg of a fruit contained 90% water one week ago. How many kg of the fruit containing 80% water are there now?

VashaNatasha [74]

90kg

It is just a simple proportional problem. If you subtract 10%, then you subtrat 10 kg.

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3 years ago

What does is the value of x in this problem

nekit [7.7K]

Answer:

the value of x is 90.....

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3 years ago

Q.16 please<br> With steps

blsea [12.9K]

The equations

$2x^2+9x-5=0$

and

$2\left(t-\dfrac12\right)^2+9\left(t-\dfrac12\right)-5=0$

are really the same, we're just setting $x=t-\dfrac12$ , or $t=x+\dfrac12$ .

$2x^2+9x-5=(2x-1)(x+5)=0\implies x=\dfrac12,x=-5$

So we get

$t=\dfrac12+\dfrac12=1,t=-5+\dfrac12=-\dfrac92$

8 0

3 years ago

I was sick for a week and never got tought this stuff can you help

Crank

1. We know that Sum of Three Angles in a Triangle is 180

⇒ m∠1 + 75 + 40 = 180

Option B is the Answer

2. We know that Sum of Three Angles in a Triangle is 180

⇒ ∠1 + 30 + 20 = 180

⇒ ∠1 + 50 = 180

⇒ ∠1 = 180 - 50

⇒∠1 = 130

3. We know that Sum of Three Angles in a Triangle is 180

⇒ ∠1 + 75 + 35 = 180

⇒ ∠1 + 110 = 180

⇒ ∠1 = 180 - 110

⇒ ∠1 = 70

4. We know that Sum of Three Angles in a Triangle is 180

⇒ ∠1 + 60 + 60 = 180

⇒ ∠1 + 120 = 180

⇒ ∠1 = 180 - 120

⇒ ∠1 = 60

5. We know that Sum of Three Angles in a Triangle is 180

We can notice that the Given Triangle is a Right Angled Triangle

We know that One Angle in Right Angled Triangle is 90

⇒ ∠2 + 90 + 30 = 180

⇒ ∠2 + 120 = 180

⇒ ∠2 = 180 - 120

⇒ ∠2 = 60

6. We know that Sum of Three Angles in a Triangle is 180

We can notice that the Given Triangle is a Right Angled Triangle

We know that One Angle in Right Angled Triangle is 90

⇒ ∠2 + 90 + 40 = 180

⇒ ∠2 + 130 = 180

⇒ ∠2 = 180 - 130

⇒ ∠2 = 50

7. We know that Sum of Three Angles in a Triangle is 180

We can notice that the Given Triangle is a Right Angled Triangle

We know that One Angle in Right Angled Triangle is 90

⇒ ∠2 + 90 + 45 = 180

⇒ ∠2 + 135 = 180

⇒ ∠2 = 180 - 135

⇒ ∠2 = 45

8. We know that the Exterior Angle of one Vertex of a Triangle is the Sum of Other two vertices interior angles.

⇒ ∠3 = 80 + 60

⇒ ∠3 = 140

9. We know that the Exterior Angle of one Vertex of a Triangle is the Sum of Other two vertices interior angles.

⇒ ∠3 = 40 + 35

⇒ ∠3 = 75

10. We know that the Exterior Angle of one Vertex of a Triangle is the Sum of Other two vertices interior angles.

⇒ ∠3 = 75 + 65

⇒ ∠3 = 140

11. We know that Sum of Three Angles in a Triangle is 180

⇒ x + x + 108 = 180

⇒ 2x + 108 = 180

⇒ 2x = 180 - 108

⇒ 2x = 72

⇒ x = 36

12. We know that Sum of Three Angles in a Triangle is 180

⇒ 60 + 60 + 3x = 180

⇒ 3x + 120 = 180

⇒ 3x = 180 - 120

⇒ 3x = 60

⇒ x = 20

13. We know that Sum of Three Angles in a Triangle is 180

We can notice that the Given Triangle is a Right Angled Triangle

We know that One Angle in Right Angled Triangle is 90

⇒ (5x + 3) + 47 + 90 = 180

⇒ 5x + 140 = 180

⇒ 5x = 180 - 140

⇒ 5x = 40

⇒ x = 8

14. We can notice that the Given Diagram is a Right angled Triangle :

⇒ ∠1 + 90 + 45 = 180

⇒ ∠1 + 135 = 180

⇒ ∠1 = 180 - 135

⇒ ∠1 = 45

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3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago