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4 years ago

Solve the linear system of equations using addition. -4x=y+3 y=-4-3

Mathematics

1 answer:

prisoha [69]4 years ago

7 0

Answer: x = 1, y = -7

Step-by-step explanation:

$-4x=y+3\\y=-7$

Then substitute y=-7 in -4x=y+3

$-4x=-7+3\\-4x=-4\\x=1$

Order: (1,-7)

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Anyone know the answer for the whole question

Anon25 [30]

Answer:

60 km

Step-by-step explanation:

a) A bearing is an angle, measured clockwise from the north direction.

(They are always given in 3 figures).

Therefore, the bearing of B from A is 090°

b) <u>Method 1</u>

Jafar drives from A to B = 2.4cm

As 1 cm represents 200 km, then 2.4 x 200 = 480 km

Kenzi drives from B to C = 1.2 + 1.5 = 2.7 cm

As 1 cm represents 200 km, then 2.7 x 200 = 540 km

To calculate how much further Kenzi drives that Jafar, subtract the distance Jafar drives from the distance Kenzi drives:

540 - 480 = 60 km

<u>Method 2</u>

The other way to do this is to calculate the difference in distance in centimeters and then convert the answer to kilometres by multiplying it by 200:

Jafar = 2.4 cm

Kenzi = 1.2 + 1.5 = 2.7 cm

Difference = 2.7 - 2.4 = 0.3 cm

Convert to km = 0.3 x 200 = 60 km

3 0

3 years ago

Consider the following recursive definition of a set S: 2,3, and 4 are members of S, so is 6 added to any member of S. Consider

neonofarm [45]

Answer:

Detailed solution is given below:

8 0

3 years ago

Pls HELP DUE IN 1 HOUR WILL GIVE BRAINLIEST THANKS AND 5 STARSSSS

adelina 88 [10]

Answer: X=-1

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6 0

3 years ago

Please help me with this Algebra I question.

horrorfan [7]

Answer:

$-\dfrac{b^2}{16a^4}$

Step-by-step explanation:

Given expression:

$\dfrac{(2a)^{-2}b^5}{-4a^2b^3}$

$\textsf{Apply the exponent rule}\quad (ab)^c=a^c \cdot b^c:$

$\implies \dfrac{2^{-2}a^{-2}b^5}{-4a^2b^3}$

Separate the variables:

$\implies \dfrac{2^{-2}}{-4}\cdot \dfrac{a^{-2}}{a^2}\cdot \dfrac{b^5}{b^3}$

$\textsf{Apply the exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies \dfrac{2^{-2}}{-4}\cdot a^{-2-2}\cdot b^{5-3}$

$\implies \dfrac{2^{-2}}{-4}\cdot a^{-4}\cdot b^{2}$

$\textsf{Apply the exponent rule}\quad a^{-n}=\dfrac{1}{a^n}:$

$\implies \dfrac{1}{-4(2^2)}\cdot \dfrac{1}{a^4}\cdot b^{2}$

$\implies -\dfrac{1}{16}\cdot \dfrac{1}{a^4}\cdot b^{2}$

Simplify:

$\implies -\dfrac{b^2}{16a^4}$

5 0

2 years ago

What is the hcf of 45 and 100

Ira Lisetskai [31]

Answer:

45 100

/ \ / \

3 × 15 2×50

/ \ / \

3× 5 2×25

/ \

5×5

45 : 3×3×5

100: 2×2×5×5

common : 3×5

HCF: 15

6 0

3 years ago