All categories

2 years ago

Raj weighs a book and rounds the weights to the nearest tenth. Does this weight round to 1.5 pounds?

Mathematics

2 answers:

slamgirl [31]2 years ago

8 0

I agree with the person above me 1.480,&:&

mel-nik [20]2 years ago

6 0

Step 1: Locate and underline the tenth place (4) and look to the right (8)

1.48

Step 2: In this case, the digit to the right 8 is 5 or above. So, we add 1 to the tenth place 4. The digit at the right 8 becomes 0, thus we get 1.5 as the answer.

In short: 1.48 rounded to the nearest tenth is 1.5.

Hope that helped

You might be interested in

9-4(2p-1) = 45 solve for p

katrin [286]

Answer: p = -4

Step-by-step explanation:

9-4(2p-1) = 45

9-8p+4 = 45

13-8p = 45

-8p = 45 - 13

-8p = 32

p = -(32/8)

p = -4

8 0

3 years ago

What are the intransitive verbs are in the sentence?

Charra [1.4K]

Verbs is the answer my friend

6 0

3 years ago

Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos

Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

7 0

3 years ago

Use the expression 5(6 + 4x) to answer the following:

marissa [1.9K]

The answer is a the answer is a

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%20%3D%20ln%205" id="TexFormula1" title="e^{2x} = ln 5" alt="e^{2x} = ln 5" align=

KiRa [710]

$e ^{2x} = ln5$

Solve for the real domain

$e ^{2x} = ln(5)$

if $f(x) =g(x),$ then $ln(f(x))= ln(g(x))$

$ln(e ^{2x} ) = ln(ln(5))$

Solve : <span> $ln(e ^{2x} ) = ln(ln(5))$
</span>
use the logarithmic definition :

$ln(e^{f(x)} ) = f(x)$

$ln(e^{2x} ) = 2x$

$2x=ln(ln(5))$

Divide both sides by 2 :

$\frac{2x}{2} = \frac{ln(ln(5))}{2}$

$x= \frac{ln(ln(5))}{2}$

hope this helps!

8 0

3 years ago