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3 years ago

PLS HELP, I need help due today Math questions, I will mark brainiest if correct

Mathematics

1 answer:

Kryger [21]3 years ago

4 0

The first answer is -1.4

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The Spanish Club has 40 members, 5 girls and 35 boys. What is the ratio of

aleksley [76]

Answer:

B. 1:7

Please give me brainliest, thank you!

Step-by-step explanation:

Information given and needed:

- 40 members

- 5 girls

- 35 boys

Solve:

girls to boys

5 to 35

(You can now simplify)

1 to 7

Answer:

B. 1:7

4 0

3 years ago

Read 2 more answers

A.45 B.30 C.90 D.180 E.40 F.55 G.130 H.50

cupoosta [38]

Answer:

If you were to make a 360 degree circle and minus the 50 thats already there you get 310 the right angle on angle 5 is 90 degrees because its a right angle so then the 310-90=220 the obtuse angle on angle 2 is 180 so 220-180=40 lastly you share those forty between angles 3 and 4 and the solution is your answer.

Glab to help!

7 0

3 years ago

y=-2x+4 Complete the missing value in the solution to the equation. (_, -2) NO work SHOWN JUST GET IT RIGHT QUICK!!!

Lesechka [4]

Answer:

x=3

Step-by-step explanation:

plug in -2 for y. then you can find the x value, which is 3

3 0

3 years ago

What is the answers to this math 3-3x6+2=

Hoochie [10]

The answer to this question is 2 because 3-3=0×6=0+2=2

3 0

3 years ago

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If z and w are complex numbers w = a+bi and z = c+di Prove algebraically that |zw| = |z||w|.

-Dominant- [34]

Answer:

To show that $\lvert zw\rvert =\lvert z\rvert \lvert w\rvert$ observe that:

Step-by-step explanation:

We have that $w=(a+bi), z=(c+di)$ . Then $zw=(c+di)(a+bi)=(ac-bd)+(cb+ad)i$ . Now, the absolute value of this product is $\lvert zw \rvert =\sqrt{(ac-bd)^{2}+(cb+ad)^{2}}=\sqrt{a^{2}c^{2}-2acbd+b^{2}d^{2}+c^{2}b^{2}+2abcd+a^{2}d^{2}}=\sqrt{a^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+a^{2}d^{2}}.$

On the other hand, $\lvert z\rvert =\sqrt{c^{2}+d^{2}}$ and $\lvert w \rvert =\sqrt{a^{2}+b^{2}}.$ Then,

$\lvert z\rvert \lvert w \rvert =\sqrt{c^{2}+b^{2}}\sqrt{a^{2}+b^{2}}=\sqrt{(c^{2}+d^{2})(a^{2}+b^{2})}=\sqrt{c^{2}a^{2}+c^{2}b^{2}+d^{2}a^{2}+d^{2}b^{2}}$

Then, it is clear that $\lvert zw \rvert =\lvert z\rvert \lvert w \rvert$ .

7 0

4 years ago