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2 years ago

8 < 4(x+1) show your work

1 answer:

5 0

8 < 4 (x + 1) Use the Distributive Property
8 < 4x + 4 Subtract 4 from both sides
4 < 4x Divide both sides by 4
1 < x

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I will give brainiest to whoever answers correctly !!

Juliette [100K]

Answer:

75.99

Step-by-step explanation:

6 0

2 years ago

FIND THE DISTANCE BETWEEN THE POINTS. ROUND THE ANSWER TO TWO DECIMAL POINTS (3,2) (8,2)

defon

Answer:5

Step-by-step explanation:

If you are looking for the distance then start counting on the first decimal and stop at the second

5 0

2 years ago

Determine whether a triangle with given side lengths will be a right triangle?

mr Goodwill [35]

Answer:

Step-by-step explanation:

Apply the Pythagorean Theorrem. Find the sum of the squares of the two shortest sides and determine whether this sum equals the square of the longest side:

#19: 5^2 + 12^2 = ? = 13^2

25 + 144 = ? = 169 This is true, so you do have a right triange in #19.

#21: 2^2 + 4^2 = ? = 7^2, or 4 + 16 = ? = 49 This is false. Not a right

triangle.

Apply this same approach (Pythagorean Theorem) to the remaining problems.

4 0

2 years ago

P(x)=-4x-2<br> Find p(a-3)

Marta_Voda [28]

Answer : -4a+10

Substitute (a-3) for x
-4(a-3)-2
-4a+12-2
-4a+10 but can also be written as -2(2a-5)

7 0

2 years ago

Question 2b only! Evaluate using the definition of the definite integral(that means using the limit of a Riemann sum

lara [203]

Answer:

Hello,

Step-by-step explanation:

We divide the interval [a b] in n equal parts.

$\Delta x=\dfrac{b-a}{n} \\\\x_i=a+\Delta x *i \ for\ i=1\ to\ n\\\\y_i=x_i^2=(a+\Delta x *i)^2=a^2+(\Delta x *i)^2+2*a*\Delta x *i\\\\\\Area\ of\ i^{th} \ rectangle=R(x_i)=\Delta x * y_i\\$

$\displaystyle \sum_{i=1}^{n} R(x_i)=\dfrac{b-a}{n}*\sum_{i=1}^{n}\ (a^2 +(\dfrac{b-a}{n})^2*i^2+2*a*\dfrac{b-a}{n}*i)\\$

$=(b-a)^2*a^2+(\dfrac{b-a}{n})^3*\dfrac{n(n+1)(2n+1)}{6} +2*a*(\dfrac{b-a}{n})^2*\dfrac{n (n+1)} {2} \\\\\displaystyle \int\limits^a_b {x^2} \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} R(x_i)\\\\=(b-a)*a^2+\dfrac{(b-a)^3 }{3} +a(b-a)^2\\\\=a^2b-a^3+\dfrac{1}{3} (b^3-3ab^2+3a^2b-a^3)+a^3+ab^2-2a^2b\\\\=\dfrac{b^3}{3}-ab^2+ab^2+a^2b+a^2b-2a^2b-\dfrac{a^3}{3} \\\\\\\boxed{\int\limits^a_b {x^2} \, dx =\dfrac{b^3}{3} -\dfrac{a^3}{3}}\\$

4 0

2 years ago