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3 years ago

Which graph represents this equation? -3x + 4y = -12

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer:

Thats a linear function. The graph should be from point (0,-3) to point (4,0)

Step-by-step explanation:

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The ratio of money in Terry's bank account to Faye's bank account was 3:7. Terry then put £220 in his account and Faye withdrew

Nadusha1986 [10]

Answer:

Terry initially had £390

Step-by-step explanation:

The ratio of the original amount is 3:7

Let the original amount Terry has be t and the original amount Faye has be f

3/7 = t/f

or simply 3f = 7t ••••••(i)

Terry putting 220 means he now has

t + 220

Faye withdrew 300

He now has f-300

These values are equal

t + 220 = f-300

f = t+220+300

f = t + 520 ••••••(ii)

Put ii into i

3(t + 520) = 7t

3t + 1560 = 7t

7t-3t = 1560

4t = 1560

t = 1560/4

t = 390

4 0

3 years ago

Choose Yes or No to indicate whether the statement is correct.

Savatey [412]

Answer:

3rd one is no. the others are all yes

4 0

3 years ago

Read 2 more answers

Joey had 26 game cards. His friend Richard gave him some more, and now Joey has 100 cards. How many cards did Richard give to Jo

Roman55 [17]

Given:

Joey had 26 game cards.

His friend Richard gave him some more, and now Joey has 100 cards.

To find:

How many cards did Richard give to Joey?

Solution:

Let x be the number of cards that Richard give to Joey.

Joey had 26 game cards.

Now, the total cards joey has $=26+x$

Joey has total 100 cards.

$26+x=100$

$x=100-26$

$x=74$

Therefore, Richard gave him 74 cards.

3 0

3 years ago

Can someone please help me

Ludmilka [50]

Is it number 21 you need help with?

4 0

3 years ago

Read 2 more answers

Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)

JulsSmile [24]

Let the given complex number

z = x + ix = $\dfrac{5-i}{3+2i}$

We have to find the standard form of complex number.

Solution:

∴ x + iy = $\dfrac{5-i}{3+2i}$

Rationalising numerator part of complex number, we get

x + iy = $\dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}$

⇒ x + iy = $\dfrac{(5-i)(3-2i)}{3^2-(2i)^2}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2}$ - $b^{2}$

⇒ x + iy = $\dfrac{15-10i-3i+2i^2}{9-4i^2}$

⇒ x + iy = $\dfrac{15-13i+2(-1)}{9-4(-1)}$ [ ∵ $i^{2} =-1$ ]

⇒ x + iy = $\dfrac{15-2-13i}{9+4}$

⇒ x + iy = $\dfrac{13-13i}{13}$

⇒ x + iy = $\dfrac{13(1-i)}{13}$

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

5 0

3 years ago