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lesantik [10]
4 years ago
9

10. A polyhedron has 5 faces and 9 edges. How many vertices does ithave?​

Mathematics
2 answers:
Ksivusya [100]4 years ago
5 0

Answer:

45

Step-by-step explanation:

san4es73 [151]4 years ago
3 0

Answer:

It has 6 vertices

Step-by-step explanation:

It is a Triangular prism

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Which statement best describes the expression (5 x 3 − 12) x 8? Subtract the product of 5 and 3 from the product of 12 and 8. Mu
Temka [501]

Answer:

Subtract 12 from the product of 5 and 3, then multiply by 8.

Explanation:

This is the only answer that accurately describes the given expression.

I hope this helps! ☺

5 0
4 years ago
Which is the solution of the quadratic equation (4y-3)^2=72 ?
max2010maxim [7]
Hello : 
<span>(4y-3)²=72 
4y-3 = </span>√72   or  4y-3 = - √72
y= (3+√72) /4   or y= (3 - √72) /4
7 0
3 years ago
In a certain chemical, the ratio of zinc to copper is 4 to 11. A jar of the chemical contains 418 grams of
olchik [2.2K]

Answer:

Step-by-step explanation:

Zinc:copper = 4:11

Amount of zinc = 4x

Amount of copper = 11x

11x = 418

x = 38 grams

Amount of zinc = 4x = 152 grams

3 0
3 years ago
5 − 2 = 25 please help me
lyudmila [28]

Answer:

Is this a true or false question? Because nothing further can be done with the equation above. If so, the equation above is false.

6 0
3 years ago
Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?
irakobra [83]

First, note that m\angle A+m\angle C=90^{\circ}. Then

m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.

Consider all options:

A.

\tan A=\dfrac{\sin A}{\sin C}

By the definition,

\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.

Now

\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.

Option A is true.

B.

\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.

Then

\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}=\dfrac{\dfrac{AB}{BC}}{\dfrac{BC}{AC}}=\dfrac{AB\cdot AC}{BC^2}\neq \dfrac{AB}{AC}.

Option B is false.

3.

\sin C = \dfrac{\cos A}{\tan C}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

Now

\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.

Option C is false.

D.

\cos A=\tan C.

By the definition,

\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.

As you can see \cos A\neq \tan C and option D is not true.

E.

\sin C = \dfrac{\cos(90^{\circ}-C)}{\tan A}.

By the definition,

\sin C=\dfrac{AB}{AC},\\ \\\cos (90^{\circ}-C)=\cos A=\dfrac{AB}{AC},\\ \\\tan A=\dfrac{BC}{AB}.

Then

\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.

This option is false.

8 0
3 years ago
Read 2 more answers
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