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adelina 88 [10]
3 years ago
10

A California condor can have a mass of 13 kilograms. A Bee hummingbird can have a mess of 1.6 grams how many times as great as t

he bee hummingbird's Mass is the California condor's mass
Mathematics
2 answers:
MariettaO [177]3 years ago
7 0
13 kg = (1000×13 ) grams 
=13000 grams

so 13000/1.6= how many times 

=8125  <span>times as great as the bee hummingbird's Mass is the California condor's mass</span>
katovenus [111]3 years ago
5 0
1 kilogram = 1,000 grams 
 a condor = 13,000 grams 
a hummingbird = 1.6 grams 

13,000 divided by 1.6 =8,125

A condor is 8,125 times bigger than a hummingbird
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you let me do the following triangle. She is transformed using a reflection which could be coordinates of dog after the reflecti
disa [49]

Answer:

The coordinates of D'O'G' after the reflection would be D=(0,0), O=(-1,2), G=(1,4). The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query about and it will be able to help you.

Step-by-step explanation:

6 0
3 years ago
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
3 years ago
Jackie created a horizontal cross section on a cylinder (soda can). What shape did she discover after making the cut?
Nesterboy [21]
It should be a circle. That would be my best guess
5 0
3 years ago
Please help!! I Will give brainliest!! Just the coordinates will be fine!!
Lerok [7]

Answer:

(2, 1) (1, 5) and (4, 3)

Step-by-step explanation:

Basically switch the x and y axes, and depending on the quadrant, switch the negative and positive signs to their appropriate ones.

hope this helps!

please heart and five-star it if u have the chance! :D

6 0
3 years ago
Use the given information to prove that RSU=TSV
RoseWind [281]

Answer:

Statement,                                                        Reason

Given segment RV = segment TU,                 Given  

m∠R = m∠T,                                                      Given

Segment RS ≅ segment ST,                            Base angles theorem

ΔRSV ≅ ΔTSU,                                                 SAS rule of congruency

Segment SV ≅ segment SU,                          CPCTC

VU ≅ UV,                                                          Reflexive property

RU = RV + VU, TV = TU + UV,                          Addition of segments

RV + VU ≅ TU + UV,                                         Addition property of equality

RU ≅ TV,                                                           Transitive property of addition

ΔRSU ≅ ΔTSV,                                                  SSS rule of congruency

Step-by-step explanation:

Where:

SAS- Side Angle Side

CPCTC -Congruent Parts of Congruent Triangles are Congruent

SSS -Side Side Side

3 0
3 years ago
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