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3 years ago

2qt= ?ml round to nearest hundredths

Mathematics

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

1982.71 mL

Step-by-step explanation:

1 qt = 946.353 mL

2 qt = 1982.71 mL

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4(t+25) = (t+50) - 4 (0.15t)

LenKa [72]

Answer:

4(t+25) = (t+50) - 4 (0.15t)

4t + 100 = t + 50 - 0.6t

4t + 100 = 50 + 0.4t

4t - 0.4t + 100 = 50 + 0.4t - 0.4t

3.6t + 100 = 50

3.6t + 100 - 100 = 50 - 100

3.6t = -50

3.6t / 3.6 = -50 / 3.6

t = - 13.9

Step-by-step explanation:

6 0

3 years ago

Plot a point on the coordinate plane to represent each of the

Juli2301 [7.4K]

(1,4) you go right 1 time, and than go up 4.

7 0

2 years ago

The two rectangular prisms are similar.

Sophie [7]

Answer:

The volume of the larger rectangular prism is $648\ cm^{3}$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z-----> the scale factor

x-----> the volume of the larger rectangular prism

y-----> the volume of the smaller rectangular prism

$z^{3}=\frac{x}{y}$

In this problem we have

$z=\frac{12}{4}=3$ ---> the scale factor is equal to the ratio of its corresponding sides

$y=24\ cm^{3}$

substitute and solve for x

$3^{3}=\frac{x}{24}$

$x=27*24=648\ cm^{3}$

6 0

3 years ago

Its -20 but how? pls helppp. ill give brainliest

MissTica

w-10=2(w+5)

w-10=2w+10

w-2w=10+10

-w=20

w=-20

4 0

2 years ago

Read 2 more answers

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.

Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

AD+DB=3

x+DB = 3

DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

$BC^{2} = DB^{2} +CD^{2}$

$BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}$

$BC^{2} =(3-x)^{2} +2$

Since given AD=BC,let us plugin BC=x in above step.

$x^{2} =(3-x)^{2} +2$

$x^{2} =9-6x+x^{2} +2$

6x=11

x= $\frac{11}{6}$

Now we know AD=x= $\frac{11}{6}$ and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

$AC^{2} =AD^{2} +DC^{2}$

$AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}$

$AC^{2} =\frac{193}{36}$

$AC=\sqrt{\frac{193}{36} }$ = 2.315cm

3 0

3 years ago