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3 years ago

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The oblique prism has a rectangular base with a width of 10 units and a length of 13 units. An oblique prism with a rectangular

base is shown. The rectangular base has a width of 10 units and a length of 13 units. The distance between the top and the bottom base is 17 units. The top base extends 8 units to the right of the bottom base. The top base extends 8 units to the right of the bottom base. What is the volume of the prism? 1,040 cubic units 1,360 cubic units 1,950 cubic units 2,210 cubic units

2 answers:

Ksivusya [100]3 years ago

8 0

Answer:

1,950 cubic units

Step-by-step explanation:

The computation of the volume of the prism is shown below:

Area of the rectangle = length × breadth

= 10 units × 13 units

= 130 units^2

Now the volume of the prism is

= Base × height

= 130 units × √(17^2 - 8^2)

= 130 units × 15 units

= 1,950 cubic units

Sladkaya [172]3 years ago

8 0

Answer:

1,950 Cubic Units

Step-by-step explanation:

Because I said so.

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Ross percentage error's is 1.69%

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4 years ago

Read 2 more answers

What expression is equivalent to [x1/4*y16]^1/2

tigry1 [53]

Answer:

The given expression is ${{{\bf x}^{\frac{\bf 1}{\bf 4}}\times y^{\bf 16}}} ^ {\frac{1}{2}} =x^{\frac {\bf 1}{\bf 8}}\times y^{\bf 4}$ .

Step-by-step explanation:

The given expression is ${{x^{\frac{1}{4}}\times y^{16}}}^\frac{1}{2}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}=\sqrt{x^{\frac{1}{4}}\times y^{16}}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}= \sqrt{x^{\frac{1}{4}}}\times \sqrt{y^{16}}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}=x^{\frac{1}{4}}^\frac{1}{2}} \times y^{4}$

${{x^{\frac{1}{4}}\times y^{16}}}^{\frac{1}{2}}= x^{\frac{1}{8}}\times y^{4}$ by using ${a^{m}} ^{n}=a^ {mn}$ property.

5 0

4 years ago

Please help! .........

stira [4]

Answer:

1) When x is 1, y is 4, and vice versa

2) When x is -6, y is 42, and when x is 7, y is 114

3) When x is -2, y is 2, and when x is 8, y is 42

Step-by-step explanation:

1)

$y=x^2-6x+9$

$y+x=5$ which can be rearranged as $y=5-x$ .

Substituting this lone y into the first equation, you get:

$5-x=x^2-6x+9$

Move everything to one side:

$x^2-5x+4=0$

Factor:

$(x-4)(x-1)=0$

$x=1, 4$

$y=4, 1$

2)

$y-30=12x$ which can be rearranged as $y=12x+30$ .

$y=x^2+11x-12$

Let's use elimination this time and subtract the first equation from the second:

$0=x^2-x-42$

Factor:

$(x-7)(x+6)=0$

$x=-6, 7$

$y= -42, 114$

3)

$y=x^2-2x+6$

$y=4x+10$

Let's set the two equations equal to each other through substitution:

$x^2-2x-6=4x+10$

Move everything to one side:

$x^2-6x-16=0$

Factor:

$(x-8)(x+2)=0$

$x=-2, 8$

$y= 2, 42$

Hope this helps!

6 0

3 years ago

Please help!! ill give BRAINLIEST!!!

Alex787 [66]

1. Similar: all angles are equivalent.

2. Similar: 2 angles and 1 side are equivalent.

3. Not enough information.

4. Not enough information.

5. Not enough information.

6. False.

7. True.

8. True.

9. True.

10. False.

6 0

4 years ago

A student club holds a meeting. The predicate M(x) denotes whether person x came to the meeting on time. The predicate O(x) refe

Novay_Z [31]

Answer:

a) $\exists \, x \in C : O(x) = 0$

b) $\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) $\{ x \in C: M(x) = 1 \} = C$

d) $\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) $\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) $\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

Step-by-step explanation:

M(x) = 1 if the person x came to the meeting, and 0 otherwise.

O(x) = 1 if the person is an officer of the club and 0 otherwise.

D(x) = 1 if the person has paid hid/her club dues and 0 otherwise.

Lets also call C the set given by the members of the club. C is the domain of the functions M, O and D.

a) If someone is not an officer, the there should be at least one value x such that O(x) = 0. This can be expressed by logic expressions this way

$\exists \, x \in C : O(x) = 0$

b) If all the officers came on time to the meeting, then for a value x such that O(x) = 1, we also have that M(x) = 1. Thus, the set of officers of the Club is contained on the set of persons which came to the meeting on time, this can be written mathematically this way:

$\{ x \in C : O(x) = 1 \} \subseteq \{ x \in C : M(x) = 1 \}$

c) If everyone was in time for the meeting, then C is equal to the set of persons who came to the meeting on time, or, equivalently, the values x such that M(x) = 1. We can write that this way:

$\{ x \in C: M(x) = 1 \} = C$

d) If everyone paid their dues or came on time to the meeting, then if we take the set of persons who came to the meeting on time and the set of the persons who paid their dues, then the union of the two sets should be the entire domain C, because otherwise there should be a person that didnt pay nor was it on time. This can be expressed logically this way:

$\{ x \in C : D(x) = 1 \} \, \cup \, \{x \in C : M(x) = 1 \} = C$

e) If at least one person paid their dues on time and came on time to the meeting, then there should be a value x on C such that M(x) and D(x) are both equal to 1. Therefore

$\exists \, x \in C : M(x) = 1 \, \wedge D(x) = 1$

f) If there is an officer who did not come on time for the meeting, then there should be a value x in C such that O(x) = 1 (x is an officer), and M(x) = 0. As a result, we have

$\exists \, x \in C : O(x) = 1 \, \wedge M(x) = 0$

I hope that works for you!

7 0

4 years ago