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3 years ago

PLZ HELP WITH THIS!!!! IM TIMED AND NEED CORRECT ANSWERS

Mathematics

1 answer:

Georgia [21]3 years ago

4 0

1. Using the formula for nth term ( = a1 * r^(n-1)

the 5th term = (-2)*(1/4)^4

= -2/256 = -1/128 (answer)

2. The 4 terms between 1 and 32 are found as follows:-

a1 = 1 and a6 = 32 so 32/1 = a6/a1

therefore 32/1 = a1r^5 / a1

so r^5 = 32 and r = 2

so 4 terms are a1r, a1r^2, a1^r^3 and a1r^4

= 2, 4, 8, 16 (answer)

3. You need the sum of 10 terms where a1 = -2 and r = 3

S10 = -2 * (3^10 - 1) / (3 - 1)

= -59058

4. Sum to infinity = a1 / (1 - r)

a1 = 1 and r = 1/2

so the answer is 1 / (1 - 1/2)

= 1 / 1/2

= 2 (answer)

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3+×/2=10 <br> Is ×/2=7<br> And does ×=14

adoni [48]

$3+\dfrac{x}{2}=10\qquad|\text{subtract 3 from both sides}\\\\\dfrac{x}{2}=7\qquad|\text{multiply both sides by 2}\\\\\boxed{x=14}$

7 0

3 years ago

Solve the following equation algebraically: 3 x squared = 375 a. x almost-equals 49 b. x almost-equals plus-or-minus 49 c. x alm

d1i1m1o1n [39]

3x^2=375
x^2=125
x=11.18

c or d, they say the same thing

4 0

3 years ago

Can someone please explain how they got this answer because the entire lesson they haven’t shown me how to graph these

I am Lyosha [343]

Answer:

Please check explanations

Step-by-step explanation:

Here, we have three types of equations and three plotted graphs

we have a quadratic equation

an exponential equation

and a linear equation

For a quadratic equation, we usually have a parabola

The first equation is quadratic and as such the first graph that is parabolic belongs to it

For an exponential equation, we usually have a graph that rises or falls before becoming flattened

The second equation represents an exponential equation so the second graph is for it

Lastly, we have a linear equation

A linear equation usually has a straight line graph

Thus, as we can see, the third graph represents the linear equation

5 0

3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 years ago

An education center offers a total of 400 math, physics and

mihalych1998 [28]

Answer: 30

Step-by-step explanation:

3+4+3=10 400/10=40 so each has 40 so math has 120 120*25=3000 for physics is 4*40 so 160 160*20=3200 9800-3000-3200=3600 3*40=120 3600/120=30

3 0

3 years ago