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lapo4ka [179]
3 years ago
11

Ella ran 6 times around her school building. then she ran 4 miles home. Her phone told her she ran 7 miles total. Which equation

represents this situation A. 6+4x=7 B. 6x+4=7 C. 6(x+4)=7
Mathematics
1 answer:
baherus [9]3 years ago
7 0
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Solve for x with a step by step explanation please<br> (3/5)x+1/4=(3/4)x-2/5
11111nata11111 [884]

Answer:

\frac{3 \times 4}{5 \times 4} x -  \frac{3 \times 5}{4 \times 5} x =  -  \frac{2 \times 4}{5 \times 4}  - \frac{1 \times 4}{5 \times 4}

\frac{12x - 15x}{20}  =  \frac{ - 8 - 4}{20}

/·20

12x-15x = -8-4

-3x= -12. /·(-1)

3x=12

x=4

Step-by-step explanation:

\frac{3}{5} x -  \frac{3}{4} x =  -  \frac{2}{5}  -  \frac{1}{4}

7 0
4 years ago
What is the domain of y = cos θ
ElenaW [278]
All real numbers / all x 
if you're asking for range as well then its -1<x<1
5 0
4 years ago
Read 2 more answers
1. What is the value of 668 + 575 + 453, rounded to
Brums [2.3K]

Answer:

E. 1,700 is your answer.

Step-by-step explanation:

What you do is you add 668 + 575 + 453 together.

668 + 575 + 453 = 1,696.

The hundreds place is the 9. Since the 9 is bigger than 4 it gets rounded up. That means the 6 in front of the 9 becomes a 7 and 9 and 6 become a zero.

E. 1,700 is your answer.

6 0
3 years ago
Roots of Quadratics 50 PTS!!!
noname [10]

The values of k for the different quadratic equation solutions are as follows

a  the equation 2x² - x + 3k = 0 has two distinct real roots

  • k < 1/24

b. the equation 5x² - 2x + (2k − 1) = 0 has equal roots

  • k = 3/5

ci the equation -x² + 3x + (k + 1) = 0 has real roots

  • k > -3.25

d the equation 3kx² - 3x + 2 = 0 has no real solutions

  • k < ± 1.633

<h3>How to solve quadratic equations to get different answers</h3>

Quadratic equations of the form ax² + bx + c = 0 is solved using the formula

-b+\frac{\sqrt{b^{2}-4ac } }{2a}     OR     -b-\frac{\sqrt{b^{2}-4ac } }{2a}

The equation b² - 4ac is called the discriminant and it is used as follows

To solve the equation and get two real roots: 2x² - x + 3k = 0

  • b² - 4ac > 0

substituting the values gives

(-1)² - 4 * 2 * 3k > 0

1 - 24k > 0

1 > 24k

divide through by coefficient of k

k < 1/24

To solve the equation and get equal roots: 5x² - 2x + (2k − 1) = 0

  • b² - 4ac = 0

substituting the values gives

(-2)² - 4 * 5 * (2k - 1) = 0

4 - 40k + 20 = 0

-40k = -24

divide through by coefficient of k

k = 3/5

To solve the equation and get real roots  -x² + 3x + (k + 1) = 0

  • b² - 4ac > 0

substituting the values gives  

(3)² - 4 * -1 * (k+1) > 0

9 + 4k + 4> 0

4k > -13

divide through by coefficient of k

k > -3.25

To solve the equation and get  no real solutions  3kx² - 3x + 2 = 0

  • b² - 4ac < 0

substituting the values gives  

(-3)² - 4 * 3k * 2 < 0

9 - 24k² > 0

9 > 24k²

divide through by coefficient of k²

k² < 24/9

k < ± 1.633

Learn more about roots of quadratic equations: brainly.com/question/26926523

#SPJ1

4 0
1 year ago
Read 2 more answers
What is 1/5 of 37% of 4,300
Tom [10]
Firstly we calculate 37% of 4300
so 37% of 4300=37×4300/100=1591
now 1/5 of 1591= 1×1591/5= 318.2
7 0
3 years ago
Read 2 more answers
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