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2 years ago

6

a candy is packaged in a box the shape of a square pyramid. How many faces does the container have? how many edges and vertices

does it have?

1 answer:

SpyIntel [72]2 years ago

7 0

Answer:

6 faces and 8 edges

Hope this was helpful!

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Express the integral as an iterated integral in six different ways, where E is the solid bounded by y=4-x^2-4z^2 and y=0

zmey [24]

Assuming you need the integral expressing the volume of $E$ , the easiest setup is to integrate with respect to $y$ first.

This is done with either

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dx\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dz\,\mathrm dx$

Thanks to symmetry, integrating with respect to either $x$ or $z$ first will be nearly identical.

First, with respect to $x$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dz\,\mathrm dy$

Next, with respec to $z$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dx\,\mathrm dy$

5 0

2 years ago

Solve the following equation simultaneously 2a-3b+10=0, 10a+6b=8

tia_tia [17]

$2a -3b +10 =0\\\\10a+ 6b -8 =0\\\\\text{Using cross multiplication method}\\\\\\ \dfrac a{(-3)(-8) - 6(10)} = \dfrac b{10(10) - 2(-8)} = \dfrac 1{ 2(6) - (-3)(10)}\\\\\\\implies \dfrac a{24-60} = \dfrac b{100+16} = \dfrac 1{12+30}\\\\\\\\\implies -\dfrac {a}{36} = \dfrac b{116} = \dfrac 1{42}\\\\\\\text{Hence}\\\\a = - \dfrac{36}{42} = -\dfrac 67 \\\\\\ b= \dfrac{116}{42} = \dfrac{58}{21}$

3 0

2 years ago

The square root of 105 would lie between which two consecutive integers?

12345 [234]

Answer:

the square root of 105 lies between 10 and 11

Step-by-step explanation:

Which perfect squares "box" 105 in? 10^2 = 100 and 11^2 = 121.

The square roots of 100 and 121 are 10 and 11.

Thus, the square root of 105 lies between 10 and 11.

8 0

2 years ago

When a number is added to six times its square, the result is twelve. Find the number.

7nadin3 [17]

Let's represent our number as x.
x is being added to 6 * x², and it equals 12
$x + 6x^{2}=12\\6x^{2}+x-12=0$
Factor:
$6x^{2}+x-12=0\\(2x+3)(3x-4)=0$
Solve for x:
$(2x+3)=0\\2x=-3\\x=\frac{-3}{2}\\\\(3x-4)=0\\3x=4\\x=\frac{4}{3}$

Therefore, x = -3/2 and 4/3.

6 0

2 years ago

Read 2 more answers

Jim is showing his work in simplifying (−8.5 + 6.1) − 1.3.

dexar [7]

9514 1404 393

Answer:

In step 4, Jim's answer is incorrect.

Step-by-step explanation:

In step 1, Jim swaps the order of addends using the commutative property of addition.

In step 2, Jim uses the distributive property to factor -1 from the final two terms. (The associative property lets Jim move parentheses.)

6.1 +(-8.5 -1.3) . . . associative property

6.1 +(-1)(8.5 +1.3) . . . distributive property

In step 3, Jim has used the properties of real numbers to form the sum of two of them.

In step 4, Jim wrote an answer of 1.1, when the answer should have been -3.7. Jim's answer is incorrect.

__

The descriptive statements about steps 2 and 4 are both true.

8 0

2 years ago