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2 years ago

Help on this pleasse.

Mathematics

1 answer:

omeli [17]2 years ago

6 0

Answer: 40 degrees

Step-by-step explanation:

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The amount of money on account it makes M for working H hours can be represented by the equation m = 29.4H what is the constant

ELEN [110]

29.4H is the equation

4 0

2 years ago

X = 7W? + 4w - 6 Y =w2 - 11w +13 X-Y Your answer should be a polynomial in standard fom.

svp [43]

Answer: $w^2-22w+19$

Step-by-step explanation:

$X=7w+4w-6\\Y=w^2-11w+13\\$

$(7w+4w-6)-(w^2-11w+13)\\7w+4w-6-w^2+11w-13\\(7w+4w+11w)+(-6-13)-w^2\\22w-19-w^2\\-w^2+22w-19$

Normally we wouldn't have our square variable be negative, so in order to get rid of that negative sign without changing the equation is by multiplying by -1

$-1(-w^2+22w-19)\\=w^2-22w+19$

4 0

3 years ago

Complete the square 2x

Elan Coil [88]

Answer:

2x squared , is basically (2 times 2) into ( x into x)

so that makes (2x)^2 = 4x^2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

2. Solve the equation using the quadratic formula. -4x2 + 40x + 16 = 0 Page 1 of 2

kumpel [21]

The solution to given quadratic equation is x = -0.3851 or x = 10.3851

Solution:

Given equation is:

$-4x^2 + 40x + 16 = 0$

To find: solve by quadratic equation formula

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text{ For} -4x^2 + 40x + 16 = 0 \text{ we have } a = -4 ; b = 40 ; c = 16$

Substituting the values of a = -4 ; b = 40 ; c = 16 in above quadratic formula we get,

$\begin{aligned}&x=\frac{-40 \pm \sqrt{40^{2}-4(-4)(16)}}{2 \times-4}\\\\&x=\frac{-40 \pm \sqrt{1600+256}}{-8}\\\\&x=\frac{-40 \pm 43.081}{-8}\\\\&x=\frac{-40+43.081}{-8} \text { or } x=\frac{-40-43.081}{-8}\\\\&x=-0.3851 \text { or } x=10.3851\end{aligned}$

Thus the solution to given quadratic equation is x = - 0.3851 or x = 10.3851

8 0

3 years ago

Pls solve and show the steps, I'll give brainliest to whoever replies first and accurately

AleksandrR [38]

3(x-1) = 2x - y

3x-3 = 2x-y

3x-2x+y = 3

x + y = 3 ----> (1)

2(x+y) = 3+3y

2x + 2y = 3 + 3y

2x + 2y - 3y = 3

2x - y = 3 -----> (2)

make equation (1) and (2) together

x + y + 2x - y = 3+3

3x = 6

x = 2

now make x = 2 in equation (1)

x + y = 3

2 + y = 3

y = 3-2

y = 1

(x,y) = (2,1)

6 0

3 years ago