0 1 2 3. What is the least number that can be made

Write "yes" or "no" to state whether or not the given side lengths would form a right triangle.

jeyben [28]

1.no 2. no 3. yes
1 does not add to 90, neither does number 2. number 3 has three values that add to 90 degrees

6 0

2 years ago

Read 2 more answers

Can someone please answer. There is one question. There's a picture thank you!

soldier1979 [14.2K]

<span>The range of Cos(x) is [-1,1]. Therefore the range of ln(Cos(x)) will be the image of [-1,1] using the natural log function. However, the domain of ln(x) is (0,infinity) and the log function is strictly increasing with vertical asymptote at "x=0". Therefore, the range of ln(Cos(x)) will be:

(- infinity, ln(1) ] = (-infinity, 0] !</span><span>so its true </span>

8 0

3 years ago

Solve this system

madam [21]

Answer:

Its the first option

Step-by-step explanation:

4 0

2 years ago

Express each ratio as a unit rate.round to the bearest tenth,if necessary. 2 note books for $0.60

FrozenT [24]

Option B

2 notebooks for $ 0.60 expressed as unit rate is $ 0.30 per notebook

<h3><u>Solution:</u></h3>

We have to express ratio as unit rate

Given that, 2 notebooks for $ 0.60

Number of notebooks = 2

Cost of 2 notebooks = $ 0.60

So we have to find cost of 1 notebook

$\begin{array}{l}{\text { cost of } 1 \text { notebook }=\frac{\text { cost of } 2 \text { notebook }}{2}} \\\\ {\text { cost of } 1 \text { notebook }=\frac{0.60}{2}=0.30}\end{array}$

So unit rate = cost of 1 notebook = $ 0.30

So 2 notebooks for $ 0.60 as unit rate is $ 0.30 per notebook

6 0

3 years ago

Read 2 more answers

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$

Now, put the value of length(2B) into equation 3 to know length (1B)

$length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm$

I really hope this helps you, have a great day.

6 0

2 years ago