Maths help please!!!!!! trigonometry

Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

ira [324]

For a system of equations to have infinitely many solutions, the equations have to be lineary dependent, (i.e. one of the equations is a multiple of the other equation). For the given equation to have infinitely many solutions, the second equation have to be a multiple of the first equation. The first equation can be rewritten as y - 2x = -5 The second equation is the first equation multiplied by 2, i.e. 2(y - 2x= -5) = 2y - 4x = -10 Therefore, the correct answer is -10 (option a).

7 0

3 years ago

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How could you double the volume of the tent just by doubling just one of its dimensions?

Vsevolod [243]

V=B*h

To double the volume , we have to double the height

then V2=B*2h=2*V

7 0

3 years ago

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The additive invers of the numer <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B7%7D%20" id="TexFormula1" title=" \fr

GuDViN [60]

Answer:

$-\frac{3}{7}$

Step-by-step explanation:

The additive inverse of a number (its "opposite") is what you add to the given number to get a sum of zero.

$\frac{3}{7} + (-\frac{3}{7}) =0$

6 0

3 years ago

3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters

Veseljchak [2.6K]

Answer:

Since the order of choice matters, we will permute the values. a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___ ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_4_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with 3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_4_ _3_ ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_4_ _3_ _2_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

Total possible words = 4 * 3 * 2

Total possible words = 24

We could've used the formula for Permutation as well

8 0

3 years ago

Use a Venn diagram to answer the question. A survey of 180 families showed that 67 had a dog; 52 had a cat; 22 had a dog and a

Mice21 [21]

Answer:

There are 13 families had a parakeet only

Step-by-step explanation:

* Lets explain the problem

- There are 180 families

- 67 families had a dog

- 52 families had a cat

- 22 families had a dog and a cat

- 70 had neither a cat nor a dog, and in addition did not have a

parakeet

- 4 had a cat, a dog, and a parakeet (4 is a part of 22 and 22 is a part

of 67 and 520

* We will explain the Venn-diagram

- A rectangle represent the total of the families

- Three intersected circles:

C represented the cat

D represented the dog

P represented the parakeet

- The common part of the three circle had 4 families

- The common part between the circle of the cat and the circle of the

dog only had 22 - 4 = 18 families

- The common part between the circle of the dog and the circle of the

parakeet only had a families

- The common part between the circle of the cat and the circle of the

parakeet only had b families

- The non-intersected part of the circle of the dog had 67 - 22 - a =

45 - a families

had dogs only

- The non-intersected part of the circle of the cat had 52 - 22 - b =

30 - b families

had cats only

- The non-intersected part of the circle of the parakeet had c families

had parakeets only

- The part out side the circles and inside the triangle has 70 families

- Look to the attached graph for more under stand

∵ The total of the families is 180

∴ The sum of all steps above is 180

∴ 45 - a + 18 + 4 + 30 - b + b + c + a + 70 = 180 ⇒ simplify

- (-a) will cancel (a) and (-b) will cancel (b)

∴ (45 + 18 + 4 + 30 + 70) + (-a + a) + (-b + b) + c = 180

∴ 167 + c = 180 ⇒ subtract 167 from both sides

∴ c = 180 - 167 = 13 families

* There are 13 families had a parakeet only

5 0

3 years ago