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2 years ago

Use your knowledge about parentheses and substitution to solve the following:

Mathematics

2 answers:

meriva2 years ago

5 0

Answer:

16. r(s)(t) =

r=2, s=3, t=4

(2)(3)(4)=

(2*3) = 6 ---> 6(4) = 24

final: 24

17. (r)(s)(t)(x)(y) =

r = 2, s = 3, t = 4, x = 5, y = 6

(2)(3)(4)(5)(6)

2*3=6

6*4=24

24*5= 120

120*6=720

final: 720

18. (7x + 2Y) =

x = 5, y = 6

rewrite---> = [7(5) + 2(6)]

[35 + 12] = 47

final: 47

hope this helpsss sorry took a while LOL

Step-by-step explanation:

q = 1; r = 2; s = 3; t = 4; x = 5; y = 6

AleksAgata [21]2 years ago

3 0

Answers:

7. 2(5 + 4) + 10 =
2 x 9 + 10
18 + 10
= 28

8. 3(6(5)) + 4 =
3 x 30 + 4
90 + 4
= 94

9. 10(5)(6) =
50 x 6
= 300

10. 5(5 + 6) =
5 x 11
= 55

11. 2(2)(3) + 4(5) =
12 + 4 x 5
12 + 20
= 32

12. 2(1) + 2(5 + 4) =
2 + 2(5 + 4)
2 + 2 x 9
2 + 18
= 20

13. 6(7(4) + 4(3)) =
6(28 + 4 x 3)
6(28 + 12)
6 x 40
= 240

14. 9(10(5)) + 10(6) =
9 x 50 + 10 x 6
9 x 50 + 60
450 + 60
= 510

15. 2(5)(5(6)) =
2 x 5 x 30
= 300

16. 2(3)(4) =
6 x 4
= 24

17. (2)(3)(4)(5)(6) =
10 x 3 x 4 x 6
30 x 4 x 6
120 x 6
= 720

18. (7(5) + 2(6)) =
35 + 2 x 6
35 + 12
= 47

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2x^2=9-3x Find the factors

d1i1m1o1n [39]

Answer:

x = -3, $\frac{3}{2}$

Step-by-step explanation:

The given quadratic equation is: $2x^2 = 9 - 3x$

This can be written as: $2x^2 + 3x - 9 = 0$

To solve a quadratic equation of the form $ax^2 + bx + c = 0$ we use the formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 2; b = 3; c = - 9

Therefore, the roots of the equation are:

$x = \frac{- 3 \pm \sqrt{9 - 4(2)(-9)}}{2(2)}$

$\implies x = \frac{-3 \pm \sqrt{81}}{4}$

$\implies x = \frac{-3 \pm 9}{4}$

We get two values of 'x', viz.,

x = $\frac{-3 + 9}{4}$ and $\frac{- 3 - 9}{4}$

$\implies x = \frac{6}{4} \hspace{5mm} \& \hspace{5mm} \frac{-12}{4}$

⇒ x = -3, 3/2

Since the factors of the quadratic equation is asked, we write it as:

(x + 3)(x - $\frac{3}{2}$ ) = 0

because, if (x - a)(x - b) are the factors of a quadratic equation, then 'a' and 'b' are its roots.

Multiply (x + 3) and (x - $\frac{3}{2}$ to see that this indeed is the given quadratic equation.

5 0

3 years ago

Describe the graph of y={1/(2x-10)}-3 compared to the graph of y=1/x

tester [92]

$\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% templates
f(x)= A( Bx+ C)+ D
\\\\
~~~~y= A( Bx+ C)+ D
\\\\
f(x)= A\sqrt{ Bx+ C}+ D
\\\\
f(x)= A(\mathbb{R})^{ Bx+ C}+ D
\\\\
f(x)= A sin\left( B x+ C \right)+ D
\\\\
--------------------$

$\bf \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\
\bullet \textit{ flips it upside-down if } A\textit{ is negative}\\
~~~~~~\textit{reflection over the x-axis}
\\\\
\bullet \textit{ flips it sideways if } B\textit{ is negative}$

$\bf ~~~~~~\textit{reflection over the y-axis}
\\\\
\bullet \textit{ horizontal shift by }\frac{ C}{ B}\\
~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\
~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by } D\\
~~~~~~if\ D\textit{ is negative, downwards}\\\\
~~~~~~if\ D\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's check these two

$\bf \stackrel{parent}{y=\cfrac{1}{x}}\qquad \qquad\qquad \qquad \stackrel{transformed}{y=\cfrac{1}{\stackrel{B}{2}x\stackrel{C}{-10}}\stackrel{D}{-3}}\\\\
-------------------------------\\\\
B=2\qquad \textit{shrinks horizontally by }\frac{1}{2}
\\\\\\
C=-10\qquad \cfrac{C}{B}=\cfrac{-10}{2}\implies -5\qquad \textit{horizontally right-shifted by }5
\\\\\\
D=-3\qquad \textit{vertically down-shifted by }3$

7 0

3 years ago

What is 2+3 and how do you solve that answer

Elden [556K]

5. Put 2 of fingers up . Then put 3 more fingers up . Count the total

3 0

3 years ago

Read 2 more answers

Is -7x=2y-6 a direct proportion?

olga_2 [115]

Yes, it is.and I have to keep typing because I need a character at least 20

8 0

3 years ago

10. 10. If AB bisects CAF, and m<EAF = 72°, then the m<BAF =

never [62]

The measure of ∠BAF is 54°.

Solution:

DF and CE are intersecting lines.

m∠EAF = 72° and AB bisects ∠CAF.

∠EAF and ∠DAC are vertically opposite angles.

Vertical angle theorem:

If two lines are intersecting, then vertically opposite angles are congruent.

∠DAC ≅ ∠EAF

m∠DAC = 72°

Sum of the adjacent angles in a straight line = 180°

m∠DAE + m∠EAF = 180°

m∠DAE + 72° = 180°

Subtract 72° from both sides.

m∠DAE = 108°

∠CAF and ∠DAE are vertically opposite angles.

⇒ m∠CAF = m∠DAE

⇒ m∠CAF = 108°

AB bisects ∠CAF means ∠CAB = ∠BAF

m∠CAB + m∠BAF = 108°

m∠BAF + m∠BAF = 108°

2 m∠BAF = 108°

Divide by 2 on both sides, we get

m∠BAF = 54°

Hence the measure of ∠BAF is 54°.

4 0

3 years ago