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3 years ago

What is the value of x? 0.9(x+1.4)−2.3+0.1x=1.6

Mathematics

2 answers:

Ivahew [28]3 years ago

7 0

this one is an easy x = 2.64

mash [69]3 years ago

5 0

Expand

Simplify 0.9x + 1.26 - 2.3 + 0.1 * x to x - 1.04

add 1.04 to both sides

simplify 1.6 + 1.04 to 2.64

Answer: x = 2.64

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dedylja [7]

Answer:

C ( 8,-6)

Because you move the points according to the rule.

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2 years ago

Read 2 more answers

Question 6

ad-work [718]

Answer:

The input for the method is a continuous function f, an interval [a, b], and the function values f(a) and f(b). The function values are of opposite sign (there is at least one zero crossing within the interval). Each iteration performs these steps: Calculate c, the midpoint of the interval, c = a + b2.

Step-by-step explanation:

trust

7 0

3 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$

3 0

3 years ago

Find the hypotenuse of each isosceles right triangle when the legs are of the given measure. 6 sqrt 2

Gemiola [76]

Isosceles right triangles have two equal sides (a and b) that are not the hypotenuse (c). And when two sides are equal, so are their opposite angles. There are only 180° degrees in any triangles, thus the right angle = 90°, so 90 left for the two equal, means that 2x=90,
x = 45°.

There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
$a = b = c \div \sqrt{2} \\ c = a\sqrt{2} \\ c = 6 \sqrt{2} \times \sqrt{2} = 6 \times 2 = 12$
Another way to do it is the longer proof of Pythagorean Theorem:
${c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\$
$c= \sqrt{({6 \sqrt{2}) }^{2} + ({6 \sqrt{2})}^{2}} \\ = \sqrt{(2 \times{(6 \sqrt{2} )}^{2} )} = \sqrt{2(36 \times 2)} \\ c = \sqrt{144} = 12$

7 0

3 years ago

5 g of margarine contains 3.6 g

Komok [63]

Answer:

59.04g of fat

Step-by-step explanation:

I) Figure out how much grams of fat is in 1g of margarine:

3.6 ÷ 5 = 0.72

II) Next, we can figure out how much fat is in 82 g of margarine:

0.72 × 82 = 59.04

One shot formula:

(3.6g ÷ 5g) × 82g = 59.04g

8 0

3 years ago