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3 years ago

Help me pleaseeeeeeeeeeeeeeee

Mathematics

1 answer:

inn [45]3 years ago

4 0

Answer:

1) n^2 – 3 n + 8 4^2 – 3*4 + 8 = 12

2) (x^2 – 4 y) / 2 (16 + 12) / 2 = 14

3) 4 * |-7 – 2| = 4 * 9 = 36 Abs -9 = 9

4) 19 – x^2 = 19 – 25 = -6 (-5)^2 = 25

5) S^2 t – 10 = -8^2 * ¾ -10 = 64 * ¾ - 10 = 38

6) 4 p^2 + 7 q^3 = 4 * 9 + 7 * (-8) = - 20 (-2)^3 = -8

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Hi, I am in desperate need of help in math. I am utterly clueless when it comes to math. Civics on the other hand. I need help w

natita [175]

First your going to use commutative property to reorder the terms , so it would be
5x^4-x-35x^3+7 , then your going to factor out x & -7 from the equation once you factor that out you should have x(5x^3-1)-7(5x^3-1) after you got that your going to factor out 5x^3-2 and get the answer (x-7)(5x^3-1) which is your final answer.

6 0

3 years ago

You build and sell toy cars for $36 each. So far, you have built 25 toy cars. How many more toy cars will you have to build and

Naily [24]

Multiple what you have already: 36x25 = 900
Subtract that from how much you need to earn:
1620-900=720
Divide how much more money you need by how much money each car is:
720/36= 20

You need 20 more cars

3 0

3 years ago

Read 2 more answers

Al factorizar el trinomio cuadrado perfecto, obtenemos el siguiente resultado: (que no se como resolver) xd algun pro que sepa r

PolarNik [594]

Answer:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

Step-by-step explanation:

<u>Trinomio Cuadrado Perfecto</u>

El producto notable llamado cuadrado de un binomio se expresa como:

$(a-b)^2=a^2-2ab+b^2$

Si se tiene un trinomio, es posible convertirlo en un cuadrado perfecto si cumple con las condiciones impuestas en la fórmula:

* El primer término es un cuadrado perfecto

* El último término es un cuadrado perfecto

* El segundo término es el doble del proudcto de los dos términos del binomio.

Tenemos la expresión:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8$

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

$\displaystyle a=\sqrt{\frac{100}{81}m^8p^{12}q^{16}z^2}$

$\displaystyle a=\frac{10}{9}m^4p^{6}q^{8}z$

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

$\displaystyle b=\sqrt{\frac{1}{49}m^2p^2z^8$

$\displaystyle b=\frac{1}{7}mpz^4$

Nos cercioramos de que el término central es 2ab:

$\displaystyle 2ab=2\frac{10}{9}m^4p^{6}q^{8}z\frac{1}{7}mpz^4$

Operando:

$\displaystyle 2ab=\frac{20}{63}m^5p^7q^8z^5$

Una vez verificado, ahora podemos decir que:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

5 0

2 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

A basketball player makes 4 out of 20 free throws. What is his shooting percentage?

DiKsa [7]

We can convert that into a fraction of 100 for the percentage.

4/20= 20/100

(multiply both by 5)

So, the basketball player makes 20% of the throws.

I hope this helps!
~kaikers

4 0

4 years ago