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3 years ago

I need help on Question 25 please ASAP

Mathematics

2 answers:

wlad13 [49]3 years ago

5 0

Answer: -13

Step-by-step explanation:

used a calculator to solve

luda_lava [24]3 years ago

3 0

Answer: - 13

USE BEDMAS to solve:

-3 · 5 + (-1)^4 · 8 - 6

= -3 · 5 + 1 · 8 - 6

= -15 + 8 - 6

= -13

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Find the value of the variables in the simplest form

Mrrafil [7]

Answer:

x = 2

Step-by-step explanation:

Using Pythagoras' identity in the right triangle

x² + x² = (2 $\sqrt{2}$ )² , that is

2x² = 8 ( divide both sides by 2 )

x² = 4 ( take the square root of both sides )

x = $\sqrt{4}$ = 2

8 0

3 years ago

Find the area of the quadrilateral with vertices A(0,0) B(2,-3) C(4,0) D(0,4).

muminat

Area of a triangle = ½ * base * height

Area of blue triangle = ½ * 4 * 4

Area of blue triangle = 8

Area of red triangle = ½ * 4 * 3

Area of red triangle = 6

TOTAL AREA = 8 + 6 = 14

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4 years ago

What is true about the net of a pyramid?

sashaice [31]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

Read 2 more answers

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

Fine the lower and upper quartiles and interquetile range.

Gnom [1K]

Answer:

Lower Quartile = 25.5

Upper Quartile = 61.5

IQR = 36

Step-by-step explanation:

Given data is:

5, 12, 17, 23, 28, 31, 37, 41, 42, 49, 54, 58, 65, 68, 73, 71

The lower quartile is the first quartile and upper quartile is the third quartile.

A median divides the data set in two equal halves.

Lower quartile also known as first quartile is the middle value in the first half.

The first half is:

5, 12, 17, 23, 28, 31, 37, 41

As the number of values is even, the quartile will be the average of two middle values

$Q_1 = \frac{23+28}{2} =\frac{51}{2}= 25.5$

Upper quartile is the average of middle two values in the second half when the number of values is even.

42, 49, 54, 58, 65, 68, 73, 71

$Q_3 = \frac{58+65}{2} = \frac{123}{2} = 61.5$

Interquartile range is the difference of third quartile and first quartile

$IQR = Q_3 - Q_1 = 61.5-25.5=36$

Hence,

Lower Quartile = 25.5

Upper Quartile = 61.5

IQR = 36

7 0

3 years ago