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Gennadij [26K]
2 years ago
15

The figure below shows a line graph and two shaded triangles that are similar:

Mathematics
1 answer:
Evgen [1.6K]2 years ago
6 0

Answer:here is something to keep in mind.

when there is a downward slope, the slope will ALWAYS have a negative slope. even if it is above the x-axis. negative slopes will tend to go from positive y-axis to negative y-axis

so personally would have to go with -4

because the x-axis is by 4 each time. and this slope is going downward.

your answer will be D

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If p = q and q = r then p = r
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Answer:

Syllogism, or C.

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3 years ago
I need help plz I do not really understand this if you can help me I’d highly appreciate
alisha [4.7K]

Answer:

Step-by-step explanation:

question no 3

In complementary angles , sum of two angles is 90 degree. so here,

angle A + angle B = 90 degree

angle A is given i.e 11 degree . substitute the value of A

11 degree + angle B= 90 degree

angle B = (90 -11) degree

angle B = 79 degree

question no 4

In supplementary angles , sum of two angles is 180 degree so here,

angle A + angle B = 180 degree

angle A is given i.e 149 degree . substitute the value of angle A

149 degree + angle B = 180 degree

angle B = (180 - 149) degree

angle B = 31 degree

8 0
2 years ago
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7 0
3 years ago
The first four ionization energies in kj/mol of a certain second- row element are 900, 1757, 14,849, and 21,007. what is the lik
ki77a [65]

Answer:<em> Berylium (Be)</em>

Step-by-step explanation:

The <em>first ionisation energy</em> is the energy required to remove one mole of the outermost electrons from one mole of gaseous atoms.

<em />

<em> Second, third, and fourth ionisation</em> energies are the energies required to remove the succesive mole of electrons of the same atoms

Successive ionisation energies are larger because, <em>once you have removed the first electron you get a positive ion</em>. Removing an electron (which is a negative charge) from a positive ion is more difficult than removing it from the neutral atom. And removing an electron from an ion with 2⁺ or  3⁺ charges is increasingly difficult.

When you find a large jump from one inoization energy to the succesive one you can predict that you are removing an electron from a closer to the nucleus orbital.

Berylium has atomic number 4. So, the number of electrons of the neutral atom is also 4. Hence, the electron configuration of beryllium is 1s² 2s².

From the given data, <em>the first four ionization energies in kJ/mol are 900, 1757, 14,849, and 21,007</em>.

From that you can calculate the following changes in the ionization energies:

  • From first to second: 1,757 kJ/mol - 900 kJ/mol = 857 kJ/mol
  • From second to third: 14,849 kJ/mol - 1,757kJ/mol = 13,092 kJ/mol
  • From third to fourth: 21,007 kJ/mol - 14,849 kJ/mol = 6,158 kJ/mol

And now you can see that there is a larger jump in the energy, a greater change, required to remove the third electron.

That is explained because the first and second electron are removed from the orbital 2s while the third electron has to be removed from the orbital 1s which is closer to the nucleus. This third electron is more strongly attracted by the nucleus and substantially  more energy is required to accomplish this work.

5 0
3 years ago
Read 2 more answers
What can you say about the nature of any other zeros of the quadratic equation , which has one complex zero? Explain your answer
svp [43]
According to the fundamental theorem of algebra, a polynomial has the number of zeros equivalent to the degree of the polynomial.

A quadratic equation is of degree 2 and hence has 2 zeros.

A complex zero is of the form a\pm bi which represents two zeros: a + bi and a - bi.

Thus, since a quadratic equation has only 2 zeros, a quadratic equation with a complex zero already have two zeros and hence will not have any other zero.
7 0
3 years ago
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