All categories

3 years ago

Hello, i need some help for questions 2-5 please

Mathematics

1 answer:

Troyanec [42]3 years ago

4 0

Answers: 

4. =9.999........

Examplantion :

5×2÷3=9.99999..........

You might be interested in

Which of these can be the graph of the equation y=mx+b

Lostsunrise [7]

See the explanation

<h2>Explanation:</h2><h2></h2>

Hello! Your figures are missing, so I could help you in a general way. The Slope-intercept form of the equation of a line is given by the form:

$y=mx+b \\ \\ \\ m:Slope \\ \\ b=y-intercept$

Case 1:

$m>0$

If the slope is positive, we will have a line as the form illustrated in figure 1. So as x increases y also increases.

Case 2:

$m$

If the slope is negative, we will have a line as the form illustrated in figure 2. So as x increases y decreases.

Case 3:

$m=0$

If the slope is zero, we will have a line as the form illustrated in figure 3. here we have a constant function.

Case 4:

m: undefined

If the slope is undefined, we will have a line as the form illustrated in figure 4.

<h2>Learn more:</h2>

Linear equations: brainly.com/question/8210863

#LearnWithBrainly

3 0

3 years ago

In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:

BabaBlast [244]

Answer

$a=0$ , $b=2$

$g_1(x)=\frac{5x}{2}$ , $g_2(x)=7-x$

Step-by-step explanation:

Given that

$\int \int Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)$

For the term $\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=\frac {2y}{5}$ and for $y$ is from $y=0$ to $y=5$ and the region $D$ , for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=\frac{5x}{2}$ to $y=5$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

$\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)$

Similarly, for the other term $\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy$ .

Limits for $x$ is from $x=0$ to $x=7-y$ and limits for $y$ is from $y=5$ to $y=7$ and the region $D$ , for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to $y$ then with respect to $x$ . So, the limits of $y$ become from $y=5$ to $y=7-x$ and limits of $x$ become from $x=0$ to $x=2$ as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

$\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)$

Hence, from equations $(i)$ , $(ii)$ and $(iii)$ , on reversing the order of integration, the required expression is

$\int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx$

$\Rightarrow \int \int Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)$

Now, compare the RHS of the equation $(iv)$ with

$\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx$

We have,

$a=0, b=2, g_1(x)=\frac{5x}{2}$ and $g_2(x)=7-x$ .

3 0

3 years ago

Can you help me with this? 7

andrezito [222]

Answer:

65º

Step-by-step explanation:

The angle of a straight line is 180º, so ∠ABD=180º and ∠ABC=(180-6x)º
The sum of the interior angles of a triangle is 180, so (x+40)º+(3x+10)º+(180-6x)º=180
We can solve from there, x+40+3x+10+180-6x=180
Combine like terms, -2x+230=180
Subtract 230, -2x=-50
Divide by -2, x=25

m∠CAB=(x+40)º=(25+40)º=65º
m∠ABC=(180-6x)º=(180-150)º=30º
m∠BCA=(3x+10)º=(75+10)º=85º

4 0

3 years ago

The line y = kx + 6 is a tangent to the graph of

Wittaler [7]

Answer:

Step-by-step explanation:

For equating the equations, you get

x2 + kx - 3x + 4 = 0 eq1

Setting the derivative of f(x) equal to 3 yields

f'(x) = 3

2x + k = 3 eq2

Now we substitute the value of k from eq2 into eq1.

x2 + x(3 - 2x) - 3x + 4 = 0

Solve for x from this equation. Then plug in the value of x into eq2 to solve for k

6 0

3 years ago

A zoo has two male lions 1/6 of the line on male lions how many lions are there at the zoo

natulia [17]

2 is 1/6 of ?
6x2=12
12 lions in total

7 0

3 years ago