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3 years ago

SOMEONE HELP ME ON THESE QUESTIONS

Mathematics

1 answer:

dem82 [27]3 years ago

4 0

1.) The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.

f = unknown number/variable [So 21 plus 5f(5 times f) equals 61]

21 + 5f = 61 [21(one-time) + 5f(number x variable) = 61(total)]

2.) Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.

j = unknown number/variable [So 17 plus 7j(7 times j) equals 87]

17 + 7j = 87

3.) n = number of calls

18 + 0.05n = 50.50

[Company charges $18 plus five cents per call(n), and the total charge was $50.50]

4.) s = the number of students

40 + 30s = 220

[Tutor charges $40 plus $30 per student(s), and the total charge was $220]

You might be interested in

304,056 trying to represent the value of this number using expanded notation we give the answer but is giving us an error so we

Rina8888 [55]

Answer:

(3 * 100,000) + (0 * 10,000) + (4 * 1000) + (0 * 100) + (5 * 10) + (6 * 1)

Step-by-step explanation:

We want to represent the value of this number using expanded notation

Mathematically, that will be;

(3 * 100,000) + (0 * 10,000) + (4 * 1000) + (0 * 100) + (5 * 10) + (6 * 1)

= 300,000 + 0 + 4000 + 0 + 50 + 6

= 304,057

7 0

3 years ago

PLEASE HELP ME If 0 < z ≤ 90 and sin(9z − 1) = cos(6z + 1), what is the value of z? z = 3 z = 4 z = 5 z = 6

Burka [1]

Answer:

z = 6

Step-by-step explanation:

We know that ...

sin(x) = cos(90 -x)

Substituting (9z-1) for x, this is ...

sin(9z -1) = cos(90 -(9z -1))

But we also are given ...

sin(9z -1) = cos(6z +1)

Equating the arguments of the cosine function, we have ...

90 -(9z -1) = 6z +1

90 = 15z . . . . . . . . . add (9z-1) to both sides

6 = z . . . . . . . . . . . . divide by 15

_____

Comment on the graph

The attached graph shows 5 solutions in the domain of interest. These come from the fact that the relation we used is actually ...

sin(x) = cos(90 +360k -x) . . . . . for any integer k

Then the above equation becomes ...

90 +360k = 15z

6 +24k = z . . . . . . . . . for any integer k

The sine and cosine functions also enjoy the relation ...

sin(x) = cos(x -90)

sin(9z -1) = cos(9z -1 -90) = cos(6z +1)

3z = 92 . . . . . equating arguments of cos( ) and adding 91-6z

z = 30 2/3

6 0

3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

6 0

3 years ago

Read 2 more answers

Show work please \sqrt(x+12)-\sqrt(2x+1)=1

Nesterboy [21]

Answer:

$x=4$

Step-by-step explanation:

Given $\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1$ , start by squaring both sides to work towards isolating $x$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2$

Recall $(a-b)^2=a^2-2ab+b^2$ and $\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}$ :

$\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1$

Isolate the radical:

$\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}$

Square both sides:

$\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2$

Expand using FOIL and $(a+b)^2=a^2+2ab+b^2$ :

$\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36$

Move everything to one side to get a quadratic:

$\displaystyle-\frac{1}{4}x^2+7x-24=0$

Solving using the quadratic formula:

A quadratic in $ax^2+bx+c$ has real solutions $\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ . In $\displaystyle-\frac{1}{4}x^2+7x-24$ , assign values:

$\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24$

Solving yields:

$\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}$