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3 years ago

7,8,9 ABC or D no explanation needed just answers and I’ll give brainly

Mathematics

2 answers:

victus00 [196]3 years ago

7 0

Answer: b, c, 4x^2y

Step-by-step explanation:

tekilochka [14]3 years ago

3 0

Answer:

7) B

8) C

9) 4x²y

Step-by-step explanation:

You might be interested in

Substitute the given values into the formula, and solve for the unknown variable.

viktelen [127]

39=1/2h = 78=h 52=(1/2 78)6+7 = 52=39+13

5 0

3 years ago

Two sets of equatic expressions are shown below in various forms: Line 1: x2 + 3x + 2 (x + 1)(x + 2) (x + 1.5)2 − 0.25 Line 2: x

kherson [118]

Answer: The correct line is

$\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.$

Step-by-step explanation: We are given the following two sets of quadratic expressions in various forms:

$\textup{Line 1: }x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25,\\\\\textup{Line 2 :}x^2+5x+6=(x+2)(x+3)=(x+2.5)^2+6.25.$

We are to select one of the lines from above that represent three equivalent expressions.

We can see that there are three different forms of a quadratic expression in each of the lines:

First one is the simplified form, second is the factorised form and third one is the vertex form.

So, to check which line is correct, we need to calculate the factorised form and the vertex form from the simplified form.

We have

$\textup{Line 1: }\\\\x^2+3x+2\\\\=x^2+2x+x+2\\\\=x(x+2)+1(x+2)\\\\=(x+1)(x+2),$

and

$x^2+3x+2\\\\=x^2+2\times x\times 1.5+(1.5)^2-(1.5)^2+2\\\\=(x+1.5)^2-2.25+2\\\\=(x+1.5)^2-0.25.$

So,

$\textup{Line 1 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2-0.25.$

Thus, Line 1 contains three equivalent expressions.

Now,

$\textup{Line 2: }\\\\x^2+5x+6\\\\=x^2+3x+2x+6\\\\=x(x+3)+2(x+3)\\\\=(x+2)(x+3),$

and

$x^2+5x+6\\\\=x^2+2\times x\times 2.5+(2.5)^2-(2.5)^2+6\\\\=(x+2.5)^2-6.25+6\\\\=(x+2.5)^2-0.25\neq (x+2.5)^2+6.25.$

So,

$\textup{Line 2 :}x^2+3x+2=(x+1)(x+2)=(x+1.5)^2+6.25.$

Thus, Line 2 does not contain three equivalent expressions.

Hence, Line 1 is correct.

7 0

4 years ago

Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x w

never [62]

<u>Answer-</u>

$\boxed{\boxed{x=90^{\circ},270^{\circ}}}$

<u>Solution-</u>

The given equation-

$\Rightarrow 2\sin^2 x-\cos^2 x-2=0$

$\sin^2 x+\cos^2 x=1\ \ \Rightarrow \cos^2 x=1-\sin^2 x$

Putting this,

$\Rightarrow 2\sin^2 x-(1-\sin^2 x)-2=0$

$\Rightarrow 2\sin^2 x-1+\sin^2 x-2=0$

$\Rightarrow 3\sin^2 x-3=0$

$\Rightarrow 3\sin^2 x=3$

$\Rightarrow \sin^2 x=1$

$\Rightarrow \sin x=\sqrt1$

$\Rightarrow \sin x=\pm 1$

$\Rightarrow \sin x=1,\ \sin x=-1$

$\Rightarrow x=\sin^{-1}(1),\ x=\sin^{-1}(-1)$

$\Rightarrow x=90^{\circ}+n360^{\circ},\ x=270^{\circ}+n360^{\circ}$

Where n=0, 1, 2, ......

As given 0° ≤ x < 360°, so putting n = 0

$\Rightarrow x=90^{\circ},\ x=270^{\circ}$

3 0

4 years ago

22. Catie is starting a babysitting business. She spent $26 to make signs to advertise. She charges an initial fee of $5 and the

Novay_Z [31]

Answer:

5+3x=26

Step-by-step explanation:

3x is the amount of money she makes for every extra hour of service.

5 is the initial amount of money she's paid

7 0

3 years ago

HELL PLEASE ASAP 25 POINTS PLEASE ASAPPPP

Andru [333]

Well I think it’s the third one but this is really advanced

7 0

3 years ago