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olga_2 [115]
3 years ago
13

Can someone help me with this

Mathematics
2 answers:
Leviafan [203]3 years ago
6 0
By illegal I'm guessing it means undefined, so when the denominator is equal to 0 because you can't divide by 0
you can either use the quadratic equation or sub the values from the multiple choice to find it. in this case it's easier to use the values given in the multiple choice
b^2-2b-8=0
b^2-2b=8
(-2)^2-2 (-2)=8
4--4=8
4+4=8
so -2 is one of the values
(-4)^2-2 (-4)=8
16+8=8
16+8=24
so -4 isnt one of the values
(4)^2-2 (4)=8
16-8=8
so 4 is one of the values
(-5)^2-2 (-5)=8
25+10=8
25+10=35
so -5 isn't one of the values
So because we know that the values are -2 and 4, and -5 and -4 aren't values, we can conclude the answer is B
sweet-ann [11.9K]3 years ago
4 0
As this is a fraction, we need to find values where the denominator is equal to 0.

b^2-2b-8=0
(b-4)(b+2)
b=4 \vee b=-2

So the answer is B.
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Answer:

140/130 > 104%

Step-by-step explanation:

When you convert 140/130 to a decimal, you get 1.07 which is 107%. 107% is greater than 104%, so 140/130 is > 104%

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Using the solubility curve, choose all of the statements that are correct.
irinina [24]

Answer:

See explanation below

Step-by-step explanation:

According to the solubility curve attached, let's analyze each sentence to see which one is correct:

<u>"Solubility of NaCl is least affected by temperature"</u>

True. as you can see in the curve, solubility of NaCl remains almost constant at any value of temperature, so in this case, it does not matter the temperature, the NaCl solubility will be the same.

<u>"completely dissolve 60 g of KBr in 100 g of water at 20 °C"</u>

True. As you can see in the graph, solubility of KBr begins at 60 g, so in 20 °C we have something dissolved there. Actually you have like 70 g dissolved there, so, this is true.

<u>"The solubility of KNO3 is the least affected by changes in temperature"</u>

False. The curve for KNO3 is exponencial, so, changes in temperature do actually affect the solubility.

<u>"100 g of KBr will dissolve completely in 100 g of water at 75 °C"</u>

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<u>"if you mix 120 g of NaClO3 in 100 g of water at 10 °C, approximately 30 g will not dissolve"</u>

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6 0
3 years ago
Calcula y comprueba las ecuaciones: plisss lo necesito alguien me puede ayudar a) 2X = 6 b) 10 + Z = 20 c) P + 9 = 11 d) 3X + 8
vesna_86 [32]

Answer:

a) x=3

b) z=10

c) P= 2

d) X=7

e) U=1

Step-by-step explanation:

Resolver una ecuación consiste en hallar los valores de la variable que hacen cierta la igualdad.

a) 2x= 6

El coeficiente es el número junto a la variable. En este caso, el coeficiente es 2. Para eliminar este número en la expresión 2x, debido a que la variable x esta multiplicada por 2, deberás dividir ambos lados de la ecuación entre 2, debido a que la operación opuesta de la multiplicación es la división.

(2x)÷2=6÷2

x= 3

Comprobar la solución de una ecuación se hace al remplazar la variable en una ecuación con el valor de la solución. La solución debería satisfacer la ecuación cuando se ingresa en esta.

En este caso:

2*3= 6

6=6

b) 10 + z= 20

En este caso se debe sumar o restar la constante que se encuentra acompañando a la variable en ambos lados de la ecuación de manera de aislar el término de la variable. En este caso:

10 - 10 + z= 20 -10

z= 10

Comprobación:

10 + z=20

10 + 10=20

20=20

c) P + 9= 11

P +9 - 9= 11 -9

P=2

Comprobación:

2 + 9= 11

11=11

d) 3X + 8 = 29

En este caso, se suma o resta la constante en ambos lados de la ecuación y luego se elimina el coeficiente de la variable mediante la división o multiplicación. Esto es:

3X + 8 - 8= 29 - 8

3X= 21

3X ÷3= 21÷3

X=7

Comprobación:

3*7 + 8=29

21+8=29

29=29

e) 2U + 8= 10

2U + 8 - 8= 10 -8

2U= 2

2U ÷2= 2÷2

U=1

Comprobación:

2*1 + 8= 10

2 + 8= 10

10=10

8 0
3 years ago
Please help me pleasee
ryzh [129]

Answer:

<em>x = y = 64° , z = 20°</em>

Step-by-step explanation:

<em>x = y</em> = \frac{180-52}{2} <em>= 64 </em>°

m∠ADC = 52° + x

m∠ADC = 52° + 64° = 116°

<em>z</em> = 180° - 116° - 44° <em>= 20</em> °

7 0
2 years ago
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