Solve 2x^2+8x-1=0 can someone type this "correctly" in photo math for me please?

You begin with $25 in a savings account and $50 in your checking account. Each week you deposit $5 into savings and $10 into che

Phantasy [73]

Answer:

The amount in checking will always be twice in any week.

Step-by-step explanation:

Let's say, $w$ represents the week, $s$ represents the saving account and $c$ represents the checking account.

Balance in saving account in any week $=s=5w+25$

Balance in checking account in any week $=c=10w+50$

Translate the statement into mathematical equation. Remember to use " = " for word "is" in the given statement.

We have to find $w$ for $2s = c$ .

Putting the values, we have:

$2(5w+25)=10w+50$

$\Rightarrow \mathrm{True\:for\:all}\:w$

Which means in each week the amount in checking is twice the amount in savings.

Best Regards!

7 0

3 years ago

A ship leaves port at 1:00 pm and travels Upper N 70 degrees Upper WN70°W at the rate of 10 mph. Another ship leaves the same p

Leni [432]

Answer:

30.4 miles

Step-by-step explanations:

Please refer to the attached image for explanations

7 0

3 years ago

Which equation shows how to find the place value relationship between the first 6 and the second 6 in 662? 60 ÷ 6 = 10 600 – 60

natita [175]

Answer:

600÷60=10

Step-by-step explanation:

six hundred is ten times sixty

6 0

3 years ago

Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2

Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Given quadratic equation:

$5x^2-6x-2=0$

Define the variables:

a = 5
b = -6
c = -2

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given quadratic equation are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

brainly.com/question/28105589

brainly.com/question/27953354

3 0

1 year ago

Read 2 more answers

Analyze the data represented in the table and select the appropriate model

Inessa05 [86]

Answer: Option B, a linear relation.

Step-by-step explanation:

Let's analyze the points in the data:

the first 3 pairs are: (0,14.70) (10,19.03) and (20, 23.36)

We can see the differences between consecutive points and see if the relation is linear:

(10,19.03) - (0,14.70) = ( 10 - 0, 19.03 - 14.70) = (10, 4.33) now we need to see the quotient: 4.33/10 = 0.433

(20, 23.36) - (10,19.03) = (20 - 10, 23.36 - 19.03) = (10, 4.33), and the quotient is the same as before.

This means that the relation is a linear relation, where the slope os 0.433 and the x-intercept is 14.70, so the equation can be written as:

y(x) = 0.433*x + 14.70

3 0

2 years ago