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3 years ago

NEED HELP ASAP 50 POINTS 20 MINS LEFT

Mathematics

1 answer:

lys-0071 [83]3 years ago

6 0

Answer:

look at the picture I have sent

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Mrs.Brown gives two tests to her class. 30% of the class passed both tests and 45% of the class passed the first test. What perc

viktelen [127]

Answer:

25%

Step-by-step explanation:

100%-30%=70%

70%-45%=25%

7 0

3 years ago

Is 8:10 equivalent to 9:12

vesna_86 [32]

Answer:

No they are not equal

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

What is the value of p in the question below. 10x9/2x4= px 5

ycow [4]

Answer:

p = 5

Step-by-step explanation:

We have that:

$\frac{x^{a}}{x^{b}} = x^{a-b}$

$\frac{x^{9}}{x^{4}} = x^{9-4} = x^{5}$

10/2 = 5

Then

$\frac{10x^{9}}{2x^{4}} = 5x^{9-4} = 5x^{5}$

Finding p:

$5x^{5} = px^{5}$

$p = \frac{5x^{5}}{x^{5}}$

$p = 5$

7 0

3 years ago

Con las cuatro cartas que se muestran en la figura se pueden formar diferentes números. Por ejemplo: 8232 o 3822. ¿Cuántos númer

densk [106]

Answer:

Se pueden formar 9 números pares.

Step-by-step explanation:

Dado que con cuatro cartas se pueden formar diferentes números, como por ejemplo 8232 o 3822, para determinar cuántos números pares de cuatro dígitos y diferentes puedes formar con estas cuatro cartas se debe realizar la siguiente tabla:

8232 - 8322 - 8223 = 2 pares 1 impar

2832 - 2823 - 2382 - 2328 - 2283 - 2238 = 4 pares 2 impares

3822 - 3282 - 3228 = 3 pares

Por lo tanto, se pueden formar 9 números pares.

5 0

3 years ago

A projectile is launched into the air. The function h(t) = –16t2 + 32t + 128 gives the height, h, in feet, of the projectile t s

Zinaida [17]

Answer:

t = 4 seconds

Step-by-step explanation:

The height of the projectile after it is launched is given by the function :

$h(t)=-16t^2+32t+128$

t is time in seconds

We need to find after how many seconds will the projectile land back on the ground. When it land, h(t)=0

So,

$-16t^2+32t+128=0$

The above is a quadratic equation. It can be solved by the formula as follows :

$t=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}$

Here, a = -16, b = 32 and c = 128

$t=\dfrac{-32\pm \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=\dfrac{-32+ \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}, \dfrac{-32\- \sqrt{(32)^2-4\times (-16)(128)} }{2\times (-16)}\\\\t=-2\ s\ \text{and}\ 4\ s$

Neglecting negative value, the projectile will land after 4 seconds.

4 0

4 years ago