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4 years ago

6

Mike drank more than half the juice in his glass.what fraction of the juice could Mike have drunk?

1 answer:

jekas [21]4 years ago

7 0

9/16 or anything else greater than 1/2 and less than 1

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Need help for this Algebra question plz!

Vikentia [17]

Answer:

2x+5y=-15

ax+bx=c

slope=-a/b so -2/5 matches up with point slope form

then to get y intercept substitite x in point slope for 0 and solve for Y to get -3

then do c/b to get y intercept, so -15/5, which equals -3.

5 0

3 years ago

Need now for study guide

likoan [24]

Answer:

17 cm.

Step-by-step explanation:

Because 12 cm + 5 cm. is 17 cm.

8 0

3 years ago

2.3 x 10^4 x 6.7 x 10^3 divide by <br> 5 x 10-^8

Brilliant_brown [7]

Answer:

0.3082

Step-by-step explanation:

4 0

3 years ago

Find the diameter of a circle with a circumference of 443 feet. Use 3.14 for pi.

marishachu [46]

Answer:

$\huge\boxed{\sf d = 141\ feet}$

Step-by-step explanation:

<u>Given Data:</u>

Circumference = C = 443 ft

<u>Required:</u>

Diameter = d = ?

<u>Formula:</u>

d = C / π

<u>Solution:</u>

d = 443 / 3.14

d = 141.0 feet

d = 141 feet

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3>

3 0

2 years ago

Read 2 more answers

let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

3 years ago