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2 years ago

Describe la situasion de esta expresión -3+4

1 answer:

6 0

Esa expresión es una simple suma, un número negativo más uno positivo ejemplo: le debes 3 dólares a tu hermano lo cual es negativo, pero tu padre te regala 4 dólares lo cual es una valor positivo porque lo tienes en tus manos, pagas los 3 que debes y te sobra un dólar de los 4 que tenias, ahora ya no debes nada y aún te sobra un dólar.

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A three‐digit number satisfies the following conditions: The digits are consecutive whole numbers in increasing order; the sum o

IRINA_888 [86]

Answer: 123
They have to add up to 3 4 5 6 or 7 so it has to be that

4 0

2 years ago

Help I over slept and only have one period to finish this test!!

Blababa [14]

Answer:

68°

Step-by-step explanation:

Okay, so m<8 is on the same plane as m<7 because line <em>q </em>is intersecting line <em>t </em>and that's supposed to add up to 180° so:

180° - 112° = 68°

So, m<7 = 68°

hope this helps:)

6 0

3 years ago

Read 2 more answers

The sum of three consecutive numbers is 33 more than the smallest.Find the number

valina [46]

Consecutive numbers. X
X+1
X+2

Add them together. 3x + 3 = 33 + x

-3 -3

3x = 30+ x
-x. -x

2x = 30

X = 15

3 0

3 years ago

Write the expression in terms of sin x and cos x only. sin(2x)+cos(3x)

BartSMP [9]

Sin2x=2sinxcosx, cos2x=1-2sin^2x
sin(2x)+cos(3x)=2sinxcosx+cos(x+2x)
cos(x+2x)=cosx(1-2sin^2(x))-sinx2sinxcosx
sin(2x)+cos(3x)=2sinxcosx(1-sinx)+cosx(1-2sin^2(x))

5 0

3 years ago

Read 2 more answers

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

8 0

3 years ago

Read 2 more answers