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True [87]
4 years ago
9

Please can somebody explain this to me

Mathematics
2 answers:
iris [78.8K]4 years ago
6 0
He needs 1/2 a pint of paint to finish his project and wants to only use one

Mekhanik [1.2K]4 years ago
5 0
He needs 1/2 a pint of paint to finish his project and wants to only use one

so which ever can he uses has to be greater than or equal to 1/2


the cans (answer choices) that are greater than 1/2 are; Cans A, C, and D
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Gretchen walked 7 / 10 miles on Saturday and 3 / 4 miles on Sunday. Without Solving do you think Gretchen walked more than a who
Lena [83]
Gretchen walked more than a whole mile on both days combined, because if on Saturday she walked 7/10 miles, she would have only needed 3/10 of a mile more to walk on Sunday to make 1 mile. On Sunday, however, she walked 3/4 miles, and using common knowledge, we know that 3/4 miles is more than 3/10. Thus, Gretchen walked more than a mile on both days combined. 
4 0
3 years ago
Solve pls brainliest!<br><br><br> I need help !!!!
antoniya [11.8K]

Answer:

m-19=58

Step-by-step explanation:

3 0
3 years ago
What is the slandered form of equation y = -3/2 x + 5
Pepsi [2]

Answer:

Ax+By=C

Step-by-step explanation:

basically the standard form of any linear equation, y=mx+b. y is the y coordinate, m is the slope, x is the x coordinate, and b is the y-intercept (where x is 0 on the line).

4 0
3 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
I need help on how to do this, please. What is the value of x?
Scorpion4ik [409]
×=167
180-12=168
× and 13 are linear pair so together they equal 180
5 0
3 years ago
Read 2 more answers
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