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2 years ago

Dana bikes 10 mi in 50 min. At this rate, how far can she bike in 90 min?

Mathematics

2 answers:

LenaWriter [7]2 years ago

8 0

The answer is 60 miles

sashaice [31]2 years ago

5 0

Answer:

the answer is 18 :))

Step-by-step explanation:

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Two equations are given below: m + 4n = 8 m = n - 2 What is the Solution ?

Sophie [7]

M+4n=8
m=n-2
we see that m=n-2
so we can subsitute 'n-2' for m in the first equation
(n-2)+4n=8
n-2+4n=8
5n-2=8
add 2 to both sides
5n=10
divide by 5
n=2
subsitute
n=2
m=n-2
m=2-2
m=0

m=0
n=2

the answer is c. 0,2

7 0

2 years ago

Leonardo wrote an equation that has an infinite number of solutions. One of the terms in Leonardo's equation is missing, as show

Tanya [424]

Answer:

The missing term is 3x

Step-by-step explanation:

Given

-(x - 1) + 5 = 2(x + 3) - _

Required

Find _

For proper identification of the parameters, represent _ with Z

So, the equation becomes

-(x - 1) + 5 = 2(x + 3) - Z

Open bracket (start from the left hand side)

-x + 1 + 5 = 2(x + 3) - Z

-x + 1 + 5 = 2x + 6 - Z

Solve like terms

-x + 6 = 2x + 6 - Z

Subtract 6 from both sides

-x + 6 - 6 = 2x + 6 - 6 - Z

-x = 2x - Z

Subtract 2x from both sides

-x - 2x = 2x - 2x - Z

-3x = -Z

Multiply both sides by -1

-3x * -1 = -Z * -1

3x = Z

Reorder

Z = 3x

Hence, the missing term is 3x

4 0

2 years ago

Read 2 more answers

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$