Which of the following units can be used to express volume? m m3 m2

Solve for t 6t=30<br><br><br><br> math

lesya [120]

$⟹\sf \: 6t = 30$

$⟹\sf \: Substitute \: the \: value \: of \: t \: as \: 30/6 \: in \: the \: above \: equation.$

$\sf \: 6t = 30 \\ \sf \: t = \frac{30}{6} \\ \sf \: t = 5$

Answer ⟶ $\boxed{\bf{t = 5}}$

4 0

2 years ago

Read 2 more answers

A 10 foot ladder is leaning up against a building. If the ladder makes a 46° angle

Gnesinka [82]

Answer:

7.193 ft

Step-by-step explanation:

|\

| \

h | \ 10 ft

| \

| 46° \

---------------------

For the 46° angle, h is the opposite leg. The hypotenuse is 10 ft. The sine is the trig ratio that relates the opposite leg to the hypotenuse.

sin A = opp/hyp

sin 46° = h/10

h = 10 sin 46°

h = 7.193

Answer: 7.193 ft

4 0

2 years ago

A certain windshield wiper is 10 inches long. The rubber blade is 6 inches long and attached at the outside end. If the wiper sw

Sloan [31]

Area of sector=fraction of sector times area of circle
fraction of sector=degrees/360
and area of circle=pir^2

so
hold a sec, we have the wiper on the outside, so we want to find the outside ring thing
do
area big sector-area small sector
or even better
(degrees/360) times (area big-areasmall)
big is 10
small is 4 (10-6=4)
areabig=10^2 times pi or 100pi
areasmall=4^2 times pi or 16pi
areabig-areasmall=100pi-16pi=84pi

then find the fraction

84pi times (150/360)=35pi square inches

4 0

3 years ago

Please help!! What Find the Error and explain why it is wrong!

Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. The real numerator in Step E should be:

$3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x$

Hence, Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

And further steps are described below:

Step D

$x^{2}-x+4 = 0$