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3 years ago

Can you solve this plz 8x9x3x5-4+7x4-2=?

Mathematics

2 answers:

amid [387]3 years ago

6 0

Answer:

7679879

Step-by-step explanation:

amm18123 years ago

5 0

Answer: 1102
Explain: 8 x 9 = 72
72 x 3 = 216
216 x 5 = 1080
1080 - 4 = 1076
1076 + 28 = 1104
1104 - 2 = 1102

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See picture^<br><br> i will give brainliest to the first person who answers

Masja [62]

Answer:

The two 50ml bottles

Step-by-step explanation:

(I will be using the $ but just change the sign to the euros)

$55/80ml= .68 per ml

45+22.5= $67.5

67.5/100 is about $.67

Therefor the two 50ml bottles is cheaper per ML and you get more product

8 0

4 years ago

Read 2 more answers

Hannah weighed 7.2 pounds at birth. Her weight was recorded at each of her infant doctor visits. The records from her appointmen

olchik [2.2K]

So If Hannah weighed 7.2 lb at birth during the six months that are shown on the table hanna puts on 8.7 during those six months so if we’re estimating what her weight will be at her one year visit

8.7(what she put on during the six months)+ 15.9 (her weight now)= 24.6

Answer: I estimate she will weigh 24.6lb at her one year visit

Hope this helps

5 0

3 years ago

Please Help! will give brainlest!

zzz [600]

Answer:

$\theta = \frac{\pi}{3} , \pi, \frac{5\pi}{ 3}$

Step-by-step explanation:

we want to solve the following trigonometric equation:

$\displaystyle - 2 { \sin}^{2} (\theta )+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

The first step of solving trigonometric equation is to rewrite the equation in terms of one trigonometric function . With Pythagorean theorem, we know that sin²x=1-cos²x . It will be helpful to rewrite the equation in terms of one trig functions. Therefore, substitute 1-cos² $\theta$ in the place of sin² $\theta$ :

$\displaystyle - 2 (1 - \cos ^{2} ) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)$

simplify:

$\displaystyle\implies - 2 (1 - \cos ^{2} (\theta) + \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies- 2 + \cos ^{2} (\theta)+ \cos( \theta) + 1 = 0, \: \theta \in [0,2\pi)\\\implies\cos ^{2} (\theta)+ \cos( \theta) -1 = 0$

Consider cos² $\theta\implies$ x. Thus,

$2 {x}^{2} + x - 1 = 0$

solving the quadratic equation yields:

${x}^{} = \frac{1}{2} \\ x = - 1$

back-substitute:

$\begin{cases} \cos( \theta) = \dfrac{1}{2} \\ \cos( \theta) = - 1 \end{cases} \theta \in[0,2\pi)$

take inverse trig in both sides

$\implies \begin{cases} \theta = \dfrac{\pi}{3} + 2n\pi\\\theta= \frac{5\pi}{3} + 2n\pi \\ \theta = \pi + 2n\pi\end{cases} \theta \in[0,2\pi) \\\\\implies\theta= \frac{\pi}{3}+\dfrac{2n\pi}{3},\theta\in [0,2\pi)\\\text{when n=0}\\ \implies \theta = \frac{\pi}{3} \\ \text{when n=1}\\ \theta= \pi\\ \text{when n=2}\\\theta=\frac{5\pi}{ 3}$