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3 years ago

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The sum S of the arithmetic sequence a, a + d, a + 2d, . . . , a + (n – 1)d is given by \small S=\frac{n}{2}\left [2a+(n-1)d \ri

ght ] . What is the sum of the integers from 1 to 100, inclusive, with the even integers between 25 and 63 omitted?

1 answer:

Vlada [557]3 years ago

6 0

Answer:

The required sum is 4214.

Step-by-step explanation:

We are given the following in the question:

An arithmetic sequence with first term a and common difference d.

The sum of n terms of A.P is given by

$S_n = \dfrac{n}{2}(2a + (n-1)d))$

We have to find:

$(1+2+3+...+98+99+100) - (26 + 28+...+60+62)$

First series:

$a = 1\\d = 1\\a_n = 100\\a + (n-1)d = 100\\\\n = 100\\\\S_{100} = \displaystyle\frac{100}{2}(2(1)+(100-1)1)\\\\S_{100} = 5050$

Second series:

$a' = 26\\d' = 2\\a'_n = 62\\a' + (n-1)d' = 62\\\\n = 19\\\\S'_{19} = \displaystyle\frac{19}{2}(2(26)+(19-1)2)\\\\S'_{19} = 836$

The require sum is:

$S_{100}-S'_{19} = 50505 - 836 = 4214$

Thus, the required sum is 4214.

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Let p = 5. Enter > , < , or = to compare the expressions. 4p4 2p−3

Y_Kistochka [10]

Answer:

The given expression at p = 5 can be written as:

$4(p^4) > 2p -3$

Step-by-step explanation:

Here, the given expressions are:

$(4p^4 ), (2p-3)$

Now, substitute the value of p = 5 in both the given expressions, we get:

$4(p^4) = 4(5^4) = 4 (625) = 2,500\\\implies 4(p^4) = 2,500$

Similarly,

$(2p-3) = 2(5) - 3 = 10 -3 = 7\\\implies 2p - 3 = 7$

So,now comparing both the values at p = 5, we get:

2,500 > 7

$\implies 4(p^4) > 2p -3$

Hence, the given expression at p = 5 can be written as:

$4(p^4) > 2p -3$

7 0

3 years ago

Please answer the question in the imageee ty

Pachacha [2.7K]

Answer:

C.

Step-by-step explanation:

6x1/2 is the same as dividing it by 2.

6/2=3

4/2=2

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3 years ago

A store sold a case of candles for $17.85 that had been marked up 10%. what was the original price?

Aleonysh [2.5K]

Step-by-step explanation:

The original price was $16.23

Explanation:

We can rewrite this problem as:

110% of what is $17.85.

"Percent" or "%" means "out of 100" or "per 100", Therefore 110% can be written as

110 x/100

.

When dealing with percents the word "of" means "times" or "to multiply".

Finally, lets call the number we are looking for "n".

Putting this altogether we can write this equation and solve for

n

while keeping the equation balanced:

<em>110/100 × n = $17.85</em>

<em>110/100 × n = $17.85 </em>

<em>110/100 × n = $17.85 100/110 × 110/100 × n = 100/110 × $17.85</em>

<em>110/100 × n = $17.85 100/110 × 110/100 × n = 100/110 × $17.85 </em>

<em>1</em><em>0</em><em>0</em><em>/</em><em>1</em><em>1</em><em>0</em><em> </em><em><</em><em>¬</em><em> </em><em>cross</em><em> out</em><em> </em><em>× </em><em>1</em><em>1</em><em>0</em><em>/</em><em>1</em><em>0</em><em>0</em><em> </em><em><</em><em>¬</em><em> </em><em>cross</em><em> out</em><em> </em><em> </em><em>× n = $ 1785/110</em>

<em>× n = $ 1785/110 </em>

n = $16.23n rounded to the nearest penny.

7 0

3 years ago

Help with geometry hw

tiny-mole [99]

QUESTION 1

Let the third side of the right angle triangle with sides $x,6$ be $l$ .

Then, from the Pythagoras Theorem;

$l^2=x^2+6^2$

$l^2=x^2+36$

Let the hypotenuse of the right angle triangle with sides 2,6 be $m$ .

Then;

$m^2=6^2+2^2$

$m^2=36+4$

$m^2=40$

Using the bigger right angle triangle,

$(x+2)^2=m^2+l^2$

$\Rightarrow (x+2)^2=40+x^2+36$

$\Rightarrow x^2+2x+4=40+x^2+36$

Group similar terms;

$x^2-x^2+2x=40+36-4$

$\Rightarrow 2x=72$

$\Rightarrow x=36$

QUESTION 2

Let the hypotenuse of the triangle with sides (x+2),4 be $k$ .

Then, $k^2=(x+2)^2+4^2$

$\Rightarrow k^2=(x+2)^2+16$

Let the hypotenuse of the right triangle with sides 2,4 be $t$ .

Then; we have $t^2=2^2+4^2$

$t^2=4+16$

$t^2=20$

We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;

$[(x+2)+2]^2=k^2+t^2$

$(x+4)^2=(x+2)^2+16+20$

$x^2+8x+16=x^2+4x+4+16+20$

$x^2-x^2+8x-4x=4+16+20-16$

$4x=24$

$x=6$

QUESTION 3

Let the hypotenuse of the triangle with sides (x+8),10 be $p$ .

Then, $p^2=(x+8)^2+10^2$

$\Rightarrow p^2=(x+8)^2+100$

Let the hypotenuse of the right triangle with sides 5,10 be $q$ .

Then; we have $q^2=5^2+10^2$

$q^2=25+100$

$q^2=125$

We apply the Pythagoras Theorem to the bigger right angle triangle to obtain;

$[(x+8)+5]^2=p^2+q^2$

$(x+13)^2=(x+8)^2+100+125$

$x^2+26x+169=x^2+16x+64+225$

$x^2-x^2+26x-16x=64+225-169$

$10x=120$

$x=12$

QUESTION 4

Let the height of the triangle be H;

Then $H^2+4^2=8^2$

$H^2=8^2-4^2$

$H^2=64-16$

$H^2=48$

Let the hypotenuse of the triangle with sides H,x be r.

Then;

$r^2=H^2+x^2$

This implies that;

$r^2=48+x^2$

We apply Pythagoras Theorem to the bigger triangle to get;

$(4+x)^2=8^2+r^2$

This implies that;

$(4+x)^2=8^2+x^2+48$

$x^2+8x+16=64+x^2+48$

$x^2-x^2+8x=64+48-16$

$8x=96$

$x=12$

QUESTION 5

Let the height of this triangle be c.

Then; $c^2+9^2=12^2$

$c^2+81=144$

$c^2=144-81$

$c^2=63$

Let the hypotenuse of the right triangle with sides x,c be j.

Then;

$j^2=c^2+x^2$

$j^2=63+x^2$

We apply Pythagoras Theorem to the bigger right triangle to obtain;

$(x+9)^2=j^2+12^2$

$(x+9)^2=63+x^2+12^2$

$x^2+18x+81=63+x^2+144$

$x^2-x^2+18x=63+144-81$

$18x=126$

$x=7$

QUESTION 6

Let the height be g.

Then;

$g^2+3^2=x^2$

$g^2=x^2-9$

Let the hypotenuse of the triangle with sides g,24, be b.

Then

$b^2=24^2+g^2$

$b^2=24^2+x^2-9$

$b^2=576+x^2-9$

$b^2=x^2+567$

We apply Pythagaoras Theorem to the bigger right triangle to get;

$x^2+b^2=27^2$

This implies that;

$x^2+x^2+567=27^2$

$x^2+x^2+567=729$

$x^2+x^2=729-567$

$2x^2=162$

$x^2=81$

Take the positive square root of both sides.

$x=\sqrt{81}$

$x=9$

QUESTION 7

Let the hypotenuse of the smaller right triangle be; n.

Then;

$n^2=x^2+2^2$

$n^2=x^2+4$

Let f be the hypotenuse of the right triangle with sides 2,(x+3), be f.

Then;

$f^2=2^2+(x+3)^2$

$f^2=4+(x+3)^2$

We apply Pythagoras Theorem to the bigger right triangle to get;

$(2x+3)^2=f^2+n^2$

$(2x+3)^2=4+(x+3)^2+x^2+4$

$4x^2+12x+9=4+x^2+6x+9+x^2+4$

$4x^2-2x^2+12x-6x=4+9+4-9$

$2x^2+6x-8=0$

$x^2+3x-4=0$

$(x-1)(x+4)=0$

$x=1,x=-4$

We are dealing with length.

$\therefore x=1$

QUESTION 8.

We apply the leg theorem to obtain;

$x(x+5)=6^2$

$x^2+5x=36$

$x^2+5x-36=0$

$(x+9)(x-4)=0$

$x=-9,x=4$

We discard the negative value;

$\therefore x=4$

QUESTION 9;

We apply the leg theorem again;

$10^2=x(x+15)$

$100=x^2+15x$

$x^2+15x-100=0$

Factor;

$(x-5)(x+20)=0$

$x=5,x=-20$

Discard the negative value;

$x=5$

QUESTION 10

According to the leg theorem;

The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the portion of the hypotenuse adjacent to that leg.

We apply the leg theorem to get;

$8^2=16x$

$64=16x$

$x=4$ units.

QUESTION 11

See attachment

Question 12

See attachment

6 0

4 years ago

(pls help I will give brainliest ) Solve for x and the missing angles

ryzh [129]

Answer:

I really hope u get the answer fingers crossed

3 0

4 years ago